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Chapter 1: Whole Numbers (Draft

)
H. Wu
Department of Mathematics #3840
Universit of California, Berkeley
Berkeley, CA 94720-3840
http://www.math.berkeley.edu/∼wu/
wu@math.berkeley.edu
Contents
1 Place Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
2 The Basic Laws of Operations . . . . . . . . . . . . . . . . . . 20
3 The Standard Algorithms . . . . . . . . . . . . . . . . . . . . 38
3.1 An addition algorithm . . . . . . . . . . . . . . . . . . . . . . 41
3.2 A subtraction algorithm . . . . . . . . . . . . . . . . . . . . . 53
3.3 A multiplication algorithm . . . . . . . . . . . . . . . . . . . . 62
3.4 Division-with-remainder . . . . . . . . . . . . . . . . . . . . . 67
3.5 The long division algorithm . . . . . . . . . . . . . . . . . . . 79
4 The Number Line and the Four Operations Revisited . . . . . 90
5 What Is a Number? . . . . . . . . . . . . . . . . . . . . . . . . 99
6 Some Comments on Estimation (not yet written) . . . . . . . 103
7 Numbers to Arbitrary Base (not yet written) . . . . . . . . . . 103
0
September 1, 2002
1
2
How to Read This Monograph
In order to learn mathematics, you will have to read arti-
cles and books in mathematics, such as this monograph. Read-
ing mathematics requires a diﬀerent skill from reading novels or
magazines, and I want to say a few words about this diﬀerence.
First of all, reading mathematics requires sustained eﬀort and
total concentration. It is a slow and painstaking process. This
monograph is not a great candidate for bedtime reading unless
insomnia is not a problem with you. Some mathematics text-
books, especially those written in the last three decades, do a lot
of “padding”, i.e., inserting long passages with little content, and
may have instilled the illusion that skipping is a good policy in
the reading of mathematics. This monograph, by contrast, says
only what needs to be said, so you will have to read every line
and try to understand every line. In fact, often you will ﬁnd
yourself struggling to understand every word in order to move
forward. On occasion, I do some chatting, but more often than
not I will be talking about straight mathematics. I have made
every eﬀort to supply you with suﬃcient details to follow the rea-
soning with ease, but I tend not to waste too many words. So
you would have to read everything carefully. In the event that I
believe something can be safely skipped on ﬁrst reading, I leave
it in indented fine-print paragraphs.
You may have gotten used to the idea that in a mathematics
book you only need to look for soundbites: understand a few
procedures and forget the rest. Not so here. This monograph tells
a coherent story, but the outline of the plot (the procedures)
is already familiar to you. It is the details in the unfolding of
the story (the reasoning) that are the focus of attention here.
Think of yourself as a detective who has to solve a murder case:
you already know going in that someone was killed, so you job is
not just to report the murder but to to ﬁnd out who did it, how
he did it, and why he did it. It is the details that matter, and
they matter a lot. Learning the details of anything is hard work.
I want to to tell tell you that most mathematicians also regard
learning mathematics as very hard work. It takes time and eﬀort,
and it may mean being stuck for a long time trying to understand
3
a particular passage. Nothing good comes easily.
It would be futile, not to say impossible, for me to anticipate
the kind of diﬃculties each of you may have in reading the mono-
graph. Experience would seem to indicate, however, that most
of you will be surprised by the emphasis in this monograph on
the importance of definitions. A (very) mistaken belief which
unfortunately has gained currency in recent years is that, in the
same way that children learn to speak whole sentences without
ﬁrst ﬁnding out the precise meaning of individual words, students
can also learn mathematics by bluﬃng their way through logi-
cal arguments and computations without ﬁnding out the precise
meaning of each concept. As a result, it is customary in schools
to teach mathematics using mathematical concepts that are only
vaguely understood. Such a belief is completely without
foundation.
Take the concept of a fraction or a decimal, for example. It is
almost never clearly deﬁned. Yet children are asked to add, multi-
ply and divide fractions and decimals without knowing what they
are or what these operations mean, and textbooks contribute to
children’s misery by never deﬁning them either. If we can get
away with this kind of mathematics education, — in the sense
that children learn all they need to learn without the beneﬁt of
clear deﬁnitions — ﬁne. But we cannot, because children are
on the whole not learning it. From the standpoint of mathemat-
ics, the ﬁrst remedy that should be tried is to explain clearly
what these concepts mean, because mathematics by its very na-
ture is a subject where everything is clearly explained. Giving
clear deﬁnitions of concepts before putting them to use has the
virtue of taking the guesswork out of learning: every step can
now be explained, and therefore more easily learned. This is the
approach taken here. If you feel uncomfortable with such an ap-
proach, can you perhaps suggest an alternative? In any case, it
is only a matter of time, and maybe a little practice, before you
get used to it. (Smokers also feel extremely uncomfortable at the
beginning of their attempt to quit smoking.) You will discover
that having clear-cut deﬁnitions is by far the better way to learn
and to teach mathematics.
At the risk of stating the obvious, I may point out that while
4
this monograph addresses serious mathematics, its exposition is
given in ordinary conversational English (or as conversational as
an ESL person can manage it). Why this is worth mentioning is
that there is at present a perception that mathematical writing
should not be couched in ordinary English. The thinking goes
roughly as follows. Because mathematics is somehow differ-
ent, it requires a different kind of writing: fewer words and
more symbols, for instance, and complete sentences are optional.
Whatever the justiﬁcation of this kind of misconception, the end
result is there for all to see: students stop using correct gammar
and syntax in their homework and exam papers, and a random
collection of symbols out of context usually passes for an expla-
nation. If we want to change such behavior among students, we
would do well to ﬁrst change this misconception about mathe-
matical writing amongst ourselves. We should never forget that
mathematics is an integral part of human culture. Doing math-
ematics is above all a normal part of human activities, and it
imposes on us the same obligations of normal human communi-
cation as any other endeavor. We must make ourselves under-
stood via the usual channels in the usual manner. The subject
matter requires greater precison of expression than a chat about
the private lives of movie stars, to be sure, but this precision is
something we try to achieve in the context of normal com-
munication, rather than in spite of it. Please keep this in mind
as you read this monograph.
5
Chapter Preview
This chapter discusses the whole numbers
0, 1, 2, 3, 4, . . .
with a view towards laying a ﬁrm foundation for the treatment of the main
topics of this monograph, namely, fractions and decimals. Notice that we
include 0 among the whole numbers. The main emphasis throughout will
not be on the well-known procedures such as the long division algorithm
— although a precise and correct statement of that algorithm is certainly
diﬃcult to ﬁnd in the literature — but on the logical reasoning that under-
lies these procedures. In mathematics, be it elementary or advanced, the
ﬁrst question you should always ask when confronted with any statement is
“Why?”. To try to ﬁnd out why something is true is a very natural human
impulse. Should you have any doubts, just observe how often pre-school
children raise this simple question with their parents each time they are in-
troduced to something new. As a teacher, your obligation is to keep alive
this sense of curiosity in a child. One way to do this is to ask yourself the
same question at all times and to ﬁnd out the answers, because it is also your
obligation to answer this question for your students. To this end, this chap-
ter will revisit a very familiar territory, — the arithmetic of whole numbers
— with the goal of explaining everything along the way. Because everything
here is familiar to you, at least as far as procedures are concerned, there is
an inherent danger that as you read this material you would put yourself
on automatic intellectual pilot and cease to think. To get you out of this
counter-productive mode, I would explicitly ask you to put yourself in the
position of a ﬁrst-time learner and to make believe that you are encountering
every topic for the ﬁrst time. I realize that it is very diﬃcult to do this be-
cause it requires a suspension of habits. Nevertheless, learning this material
is so crucial for the understanding of the rest of the monograph that I must
ask you to please try your best. Once you get the hang of it, you would not
only acquire an enhanced appreciation of the marvelous qualities of many
things you have always taken for granted, but also ﬁnd yourself in a much
better position to learn the materials in the later chapters on decimals and
fractions.
1 Place Value 6
1 Place Value
One cannot understand the arithmetic of whole numbers without a basic
understanding of our numeral system, the so-called Hindu-Arabic numeral
system.
1
It became the universal numeral system in the West circa 1600. This
chapter discusses only the whole numbers 0, 1, 2, 3, . . . , and the discussion
of this numeral system will continue in Chapters 4 and 5. We are here mainly
concerned with the fact that a symbol such as 2 in the number 2541 stands
not for 2 but 2000. In fact, 2541 means
2000 + 500 + 40 + 1 ,
i.e., two thousand ﬁve hundred and forty-one. As is well-known, the ten
symbols 0, 1, 2, . . . , 9 are called digits. The digit “2” in 2541, being in the
fourth place (position) from the right, stands not for 2 but 2000, i.e., two
thousand, as mentioned above. Similarly, the digit “5” being in the third
place from the right stands not for 5, but for 500, i.e., ﬁve hundred, the
“4” in the second place from the right stands for 40, and “1” being in the
right-most place means just 1. Similarly, 64738 denotes
60000 + 4000 + 700 + 30 + 8
and 6001 denotes
6000 + 1,
and so on. These examples illustrate a fundamental and fruitful idea of
representing numbers, no matter how large, by the use of only ten symbols
0, 1, 2, . . . , 8, 9, and by the use of their place in the number symbol to
represent diﬀerent magnitudes (sizes). Thus the “5” in 125 represents a
completely diﬀerent order of magnitude from the “5” in 2541, namely, 5 and
500 respectively. The term place value means that the value (magnitude,
size) of each digit depends on its place in the numeral symbol.
1
This is most likely a misnomer. The Chinese had a decimal numeral system since the
ﬁrst available record of writing dating back to at least 1000 B.C., and the rod numeral
system (also called the counting board numeral system) which has been ﬁrmly in place no
later than 200 A.D. is identical to the Hindu-Arabic system except for the ten symbols 0,
1, 2, . . . themselves. Moreover, negative numbers and decimal fractions (see Chapter 4)
have been part of the rod numeral system from the beginning. Because of the long history
of contact between the Indians and the Chinese, it may be diﬃcult to separate what is
Indian and what is Chinese in the Hindu numeral system. The much needed research has
not yet been done.
1 Place Value 7
In case the idea of place value has become too commonplace to strike you
as noteworthy, let us look at a diﬀerent numeral system for comparison: in
Roman numerals,
2
the number 33 is represented by XXXIII. Observe then
that the three “X’s” are in three diﬀerent places, yet each and every one
of them stands for 10, not 100 or 1000. Just 10. Similarly, the three “I’s”
occupy diﬀerent places too, but they all stand for 1, period. Contrast this
with the numeral 111 in our numeral system: the ﬁrst 1 on the left stands
for 100, the second stands for 10, and only the third stands for 1 itself. You
see the diﬀerence.
We have used the concept of addition to explain place value (e.g., 125 =
100 +20 +5). We could have pretended that you didn’t know what it means
to add whole numbers and give a precise deﬁnition, but that would be too
pedantic. People seem to have no problem with understanding this concept.
But we will carefully and precisely deﬁne the other three arithmetic opera-
tions in this and the the next two sections, i.e., subtraction, multiplication,
and division.
So far we have not brought out the signiﬁcance of the fact that only ten
symbols 0, 1, 2, . . . , 8, 9 — instead of twenty or sixty, say — are used to
denote any number, no matter how large. We now ﬁll in this gap. Like
place value, the fact that only ten symbols are used is easy to overlook due
to constant usage. It should be pointed out, therefore, that the great virtue
of the Hindu-Arabic system lies precisely in the systematic and combined
application of both ideas — place value and a ﬁxed small number of symbols
(ten, to be exact) — to generate all the numbers. No other numeral system of
the world (except the rod numeral system of China, see footnote 1) has ever
attained the same degree of symbolic economy. The Babylonians in the B.C.
era, for instance, used a numeral system that used place value only partially,
and the symbolic representation of some numbers became unwieldy. Since
we have already mentioned the Roman numerals, let us use it to illustrate
how complications arise from trying to cope with large numbers when place
value is not systematically applied. The symbol with the largest numerical
value in the Roman system is M, which denotes a thousand, 1000. In order
to write a million, which is a thousand thousand, one would have to write
2
It is well to point out that even in the Roman numeral system, there is a partial place
value at work. For instance “VI” is 6 while ”IV” is 4.
1 Place Value 8
MMM . . . MM (a thousand times). The Romans were spared this drudgery
apparently because they never had to deal with a large number of this size.
A latter day ad hoc convention to improve on the Roman system is to add
a bar above each symbol to increase its value a thousand times. Thus X
would denote not 10 but 10000, and M would denote a million. Even such
a desperate eﬀort cannot save this numeral system from certain disaster,
however, because it would still be too clumsy. For instance, the simplest way
to write 388999 is
CCCLXXXV IIICMXCIX.
It is conceivable that had Roman numerals been adopted as the universal
numeral system in the modern era, someone would have tried to introduce
symbols for “ten thousand” and “hundred thousand” to simplify the writing
of 388999, but if so, what about writing 60,845,279,037? Presumably, more
symbols would yet be introduced. Would you want to waste your time learn-
ing how to navigate in such a system?
To truly understand place value, we must review the process of counting
from 0, 1, 2, . . . onward in order to see how the whole numbers develop in
the Hindu-Arabic numeral system. After 0, 1, . . . , 8, 9, we have used up
all possible symbols by allowing ourselves only one place (position), the so-
called ones place. To generate the next set of numbers without adding more
symbols, the only option available is to put the same synbols in an additional
place, the so-called tens place, which by convention is to the left of the ones
place. (Keep in mind that this is no more than a convention.) In our minds,
we may think of 0, 1, 2, . . . , 8, 9 as 00, 01, 02, . . . , 08, 09; in other words
the single-digit numbers may be thought of as two-digit ones with 0 in the
tens place.
3
From this point of view, the next number after 9 is naturally
10, i.e., since 0 has already been used in the tens place, we replace 0 by its
successor 1. Thus we change the 0 of 09 to 1, and start the counting in the
ones place all over again with 0. The numbers after 09 are then 10, 11, 12,
. . . , 18, 19. The same outlook then guides us to write the number after 19
as 20, because after having used up all the digits in the ones place with the
tens place occupied by 1, it is natural to increase the latter from 1 to 2 and
start the counting in the ones place all over again with 0. Thus the next
numbers are 20, 21, 22, etc. Continuing this way, we get to 97, 98, 99. At
3
There is a further discussion of this issue of having 0’s to the left of a number at the
end of this section.
1 Place Value 9
this point, we have used up all ten digits in both the tens place and the ones
place. To proceed further, and without introducing more symbols, we will
have to make use of another place to the left of the tens place, the so-called
hundreds place (the third place from the right). Thinking of 99 again as
099, the same consideration then dictates that the next number is 100, to be
followed by 101, 102, etc. Thus we come to 109, and the next is 110, followed
in succession by 111, 112, 113, . . . , 198, 199. After that come 200, 201, 202,
etc., for exactly the same reason.
We pause to note that, between the numbers 0 and 100, if we skip count
by 10’s, then we have 00, 10, 20, . . . , 80, 90, 100. Thus in ten steps of 10’s,
we go from 0 to 100.
By the time we reach 999, again we have used up all ten digits in three
places so that the next number will have to make use of a fourth place, the
so-called thousands place (fourth place from the right). It will have four
digits and it has to be 1000 because 999 can be thought of as 0999 and we
naturally increase the 0 of 0999 to 1 and start counting all over again in the
ones, tens, and hundreds places. In general, whenever we reach the number
99 . . . 9 (n times for any nonzero whole number n), the next number must
be 100 . . . 0 (n zeros). As before, we make the following observation: if we
count from 0 to 100 . . . 0 (n zeros) in steps of 100 . . . 0 (n − 1 zeros) for any
nonzero whole number n, then we have
0 00 . . . 0

n−1
, 1 00 . . . 0

n−1
, 2 00 . . . 0

n−1
, . . . , 9 00 . . . 0

n−1
, 1 000 . . . 0

n
.
In other words, in ten steps of 100 . . . 0 (n −1 zeros), we get to 1000 . . . 0 (n
zeros) from 0.
Thus we can make three observations about the way counting is done in
the Hindu-Arabic numeral system: for any nonzero whole number n,
(i) an n digit number precedes any number with more than n
digits,
(ii) given two n digit numbers a and b, if the n-th digit (from the
right) of a precedes the n-th digit of b, then a precedes b, and
(iii) the sum of ten 1 00 . . . 0

n−1
’s is 1 00 . . . 0

n
.
In view of the previous comments about adding 0’s in front of a number,
we should add the following clariﬁcation: a number is said to be an n digit
1 Place Value 10
number if, counting from the right, the last nonzero digit is in the n-th place.
For example, 0050000 is a 5-digit number, and 1234 is a 4-digit number. As
illustrations of (i)–(iii): 987 precedes 1123, 65739 prcedes 70001, and
10000 + 10000 + + 10000

10
= 100000.
This is the right place to review and make precise the common notion of
“bigger than”. Formally, for two whole numbers a and b, we deﬁne b to be
bigger than a (or what is the same, a to be smaller than b) if, in the method
of counting described above, a comes before b. In symbols:
a < b or b > a.
Note that one sometimes says greater than in place of bigger than, and less
than in place of smaller than. If we want to allow for the possibility that b
is bigger than or equal to a, then we write:
a ≤ b or b ≥ a.
Thus 13 ≤ 13 and 7 ≥ 7, but 7 < 13 and 9356 < 11121, etc. In particular,
it is always the case that
if n is a nonzero whole number, then n > 0.
It follows from observations (i) and (ii) that
(iv) if a, b are whole numbers and b has more digits than a, then
a < b, and
(v) if a, b are two whole numbers with n digits and the n-th digit
of a is smaller than the n-th digit of b, then a < b.
For example, 872 < 1304, 100002 > 99817, 803429 < 911104, etc.
We next turn our attention to the phenomenon of “too many zeros”.
Consider a moderate-size number such as the number of seconds in a 365-
day year: 31536000. As we know, this means:
30000000 + 1000000 + 500000 + 30000 + 6000.
1 Place Value 11
I hope you are already weary of reading and trying to keep track of so many
zeros. It could be much worse, of course. For example, the age of the
universe has now been established (by observations of the Hubble telescope)
to be approximately 14000000000 (9 zeros) years. Back in the third century
B.C., Archimedes estimated that the number of grains that can be packed
into a ball the size of the then-known universe was
1, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000,
000, 000, 000, 000, 000, 000, 000, 000, 000
(63 zeros). Clearly, we must devise a shorthand notation to deal with these
zeros. (In case you haven’t noticed, symbolic notation has always arisen from
the need for better human communication. The Hindu-Arabic numeral sys-
tem — as you have seen — is a good example.)
We need to introduce a new notation and to review an old one. First the
new notation: we write 10
1
for 10, 10
2
for 100, 10
3
for 1000, 10
4
for 10000,
10
5
for 100000, and in general, write
10
n
for 1 00 . . . 0

n
(1)
where n is any whole number > 0. For notation consistency, we also write
10
0
for 1.
(You may think of 10
0
as “zero number of 0’s”.) The number n in 10
n
is
called the power or exponent of 10
n
, and 10
n
is read as “10 to the n-th power”.
Next, we recall the deﬁnition of multiplication among whole numbers as a
shorthand notation for repeated addition. In other words, 35 by deﬁnition
5 + 5 + 5 , and in general, if m, k are whole numbers, the deﬁnition of mk
(the accepted abbreviation of mk) is:
mk =



0 if m = 0
k + k + + k

m
if m ,= 0 (2)
Sometimes we refer to mk as the product of m and k, and call m and k the
factors of mk. We call attention to the fact that mk, the multiplication of k
by m, is a shorthand notation for adding k to itself m times, no more and
no less. Please be sure to impress this fact on your students. (See Exercise
1.8.)
1 Place Value 12
Activity: Consider the following introduction to multiplication
taken from a third grade textbook (the text has the goal of mak-
ing sure that at the end of the third grade, students know the
multiplication table of numbers up to 10):
Look at the 3 strips of stickers shown on the right
(there is a picture of three strips of stickers). There
are 5 stickers on each strip. How can you ﬁnd the
number of stickers there are in all?
You can ﬁnd the total number in different ways.
You can write an You can write a
addition sentence. multiplication
5 + 5 + 5 = 15 sentence
3 5 = 15
Think: 3 groups of 5 Read: Three times
= 15. 5 equals 15.
Answer: 15 stickers.
Do you think this is an ideal way to convey to third graders what
multiplication means?
The deﬁnition of multiplication in (2) lends itself to an easy pictorial
representation. For example, 3 5 , which is 5 +5 +5 , can be represented
by three rows of ﬁve dots:
• • • • •
• • • • •
• • • • •
Similarly, mk (=
m

k
1 Place Value 13
There should be no mistaking the fact that the deﬁnition in (2) means
exactly that mk is adding k to itself m times, and NOT adding m to itself
k times. We are taking nothing about multiplication for granted, so that if
we wish to say mk actually also equals adding m to itself k times, we would
have to explain why. This will be done later in ¸2, but for now we don’t need
this distraction.
We now put the new information to use: if n is a nonzero whole number,
then
2 10
n
= 1 00 . . . 0

n
.
We can now revisit 31536000 and rewrite it as
31536000 = (3 10
7
) + (1 10
6
) + (5 10
5
) + (3 10
4
) + (6 10
3
).
(We recall the convention concerning parentheses: do the computations within
the parentheses ﬁrst.) Similarly, the age of the universe is approximately
14, 000, 000, 000 = 10, 000, 000, 000 +4, 000, 000, 000 = (1 10
10
) +(4 10
9
),
and Archimedes’ number of grains of sand is simply 10
63
.
In general, a whole number such as 830159 can now be written as
830159 = (8 10
5
) + (3 10
4
) + (1 10
2
) + (5 10
1
) + (9 10
0
) .
Such an expression is sometimes referred to as the expanded form of a number,
in this case, 830159. Another example is
2070040 = (2 10
6
) + (7 10
4
) + (4 10
1
).
One advantage of writing a number in its expanded form is that it instantly
reveals the true value of a given digit in the number. For example, the 3
1 Place Value 14
of 830159 is given on the right as 3 10
4
, which immediately signals that
it stands for 30000 and not 3. The notation also gives a clear and precise
location (place) of a digit in the number symbol: the exponent, say k, of
10 in the expanded form of a number indicates precisely that the associated
digit is the (k +1)-th one from the right. Thus the expression 310
4
tells us
that 3 is in the ﬁfth place (from the right) of 830159, and 2 10
6
indicates
the position of 2 as the seventh digit from the right of 2070040. The clear
location of a digit in the expanded form of a number will turn out to be very
helpful in understanding all the arithmetic algorithms in ¸3.
There is one aspect of the expanded form that may trouble you: why use
the cumbersome notation of 5 10
1
and 9 10
0
in the expanded form of
830159 instead of just 5 10 and 9 ? The answer is: it all depends on what
we want. When absolute conceptual clarity is called for, we will use 5 10
1
and 910
0
, such as when we discuss the complete expanded form of a number
later on in ¸3 of Chapter 4. On the other hand, the less rigid notation of
5 10 and 9 is suﬃcient for ordinary purposes, so usually we just write:
830159 = (8 10
5
) + (3 10
4
) + (1 10
2
) + (5 10) + 9 .
It remains for us to tie up some loose ends in the foregoing discussion.
The ﬁrst one is that some of you may have encountered another deﬁnition of
10
n
for a nonzero whole number n as
10 10 10

n
,
whereas we have deﬁned 10
n
in (1) as
10
n
= 1 00 . . . 0

n
.
We need to clarify this situation by proving that these two numbers are the
same, i.e., we have to prove:
10
n
= 10 10 10

n
(3)
for all whole numbers n > 0.
1 Place Value 15
Activity: Prove that (3) is true for n = 1, 2, 3.
You may think that (3) is obvious, but it could only be because you are
used to thinking of n as a small number, say n = 1, 2, 3. But what about
n = 123457321009? Can you even make sense of (3) in that case? Thinking
about these questions will perhaps convince you that we should spend some
time ﬁnding out why (3) is true for all values of n.
A second loose end we should tie up is a fact that many of you probably
take for granted, namely, that for any number, e.g., 2,133,070, multiplying
it by 100,000 (say) results in a number that is obtained from 2,133,070 by
tagging the 5 zeros of 100,000 to the right of 2,133,070. That is,
2, 133, 070 100, 000 = 213, 307, 000, 000
The last loose end we wish to address is the issue of the implicit zeros in
front of any whole number, i.e., 0023 is the same as 23.
Let us ﬁrst explain why (3) above is true. The reason is twofold:
(a) It is true for n = 1, and
(b) if we introduce the temporary notation that
10[n] = 10 10 10

n
,
then both symbols 10
n
(= 1 00 . . . 0

n
) and 10[n] are “built up” in exactly the
same way, in the sense that for any nonzero whole number n,
10
n+1
= 10 10
n
and 10[n + 1] = 10 10[n]. (¸)
Note that using the temporary notation, (3) may be restated as
10
n
= 10[n] for all whole numbers n > 0.
Let us ﬁrst make use of (a) and (b) to prove this before we prove (a) and (b)
themselves.
For n = 1, it is quite obvious that both 10
1
and 10[1] are equal to 10
and therefore equal. Next, why is 10
2
= 10[2]? This is because
10
2
= 10 10
1
(by ﬁrst equality of (¸) with n = 1)
= 10 10[1] (by 10
1
= 10[1])
= 10[2] (by second equality of (¸) with n = 1)
1 Place Value 16
Next we prove 10
3
= 10[3]:
10
3
= 10 10
2
(by ﬁrst equality of (¸) with n = 2)
= 10 10[2] (by 10
2
= 10[2])
= 10[3] (by second equality of (¸) with n = 2)
Next we prove 10
4
= 10[4]:
10
4
= 10 10
3
(by ﬁrst equality of (¸) with n = 3)
= 10 10[3] (by 10
3
= 10[3])
= 10[4] (by second equality of (¸) with n = 3)
At this point, it is clear that the proof of 10
5
= 10[5] will follow exactly the
same pattern, and then 10
6
= 10[6], 10
7
= 10[7], etc. So we see that (3)
must be true in general for all n > 0.
We now take care of (a) and (b). Clearly (a) does not need any proof.
In order to prove (b), it is enough to prove (¸). First recall observation (iii)
made during the earlier discussion of counting in the Hindu-Arabic numeral
system, namely, for any whole number n,
the sum of ten 1 00 . . . 0

n
’s is 1 00 . . . 0

n+1
.
This can be expressed in the present notation as:
10 10
n
=
10

1 00 . . . 0

n
+1 00 . . . 0

n
+ + 1 00 . . . 0

n
(by deﬁnition of )
= 1 00 . . . 0

n+1
= 10
n+1
This proves one half of (¸), and the other half is easier:
10[n + 1] =
n+1

m+n
= 10
m+n
So we see that (4) is true.
4
It is a point well worth making that, without (3), it is not at all clear why
(4) should be true. Remember, 10
m
is by deﬁnition the number 100 . . . 00 (m
zeros), so the left side of (4) means (by the deﬁnition of multiplication in (2))
adding 100 . . . 0 (n zeros) to itself 100 . . . 00 (m zeros) times. If n is a big
number, such as 21334658, and m = 68739, can you be absolutely certain
that adding 100 . . . 0 (21334658 zeros) to itself 100 . . . 00 (68739 zeros) times
would result in the number
1000 . . . 00 ((68739 + 21334658) zeros) ?
So the key point of the proof of (4) is that, by using (3), the evaluation of
the product 10
m
10
n
is reduced to the counting of the number of copies
of 10’s, which we could do without any diﬃculty.
4
But see the discussion of the associative law of multiplication in ¸2.
1 Place Value 18
We are now in a position to explain the fact that multiplying 2,133,070
by 100,000 gives a number which can be obtained from 2,133,070 by adding
the 5 zeros of 100,000 to the right of 2,133,070. That is,
2, 133, 070 100, 000 = 213, 307, 000, 000
Note that, appearance notwithstanding, this fact is again not obvious! Ac-
cording to (2), the left side means adding 100,000 to itself 2,133,070 times.
The last 5 digits of the resulting number will be 0’s, to be sure, but why are
the beginning digits exactly 2133070 ? The reason is:
2, 133, 070 100, 000 = ¦ (2 10
6
) + (1 10
5
) + (3 10
4
)
+(3 10
3
) + (7 10
1
) ¦ 10
5
= (2 10
6
10
5
) + (1 10
5
10
5
) +
(3 10
4
10
5
) + (3 10
3
10
5
) +
(7 10
1
10
5
)
= (2 10
11
) + (1 10
10
) + (3 10
9
) +
(3 10
8
) + (7 10
6
) (using (4))
= 213, 307, 000, 000
The same reasoning of course allows us to write 213,307,000,000 more simply
as 213, 307 10
6
. This is, incidentally, our ﬁrst substantive application of
the expanded form of a number,
5
but it will hardly be the last.
In a similar way, we can now write the number of seconds in a year
as 31, 536 10
3
, and the age of the universe as 14 10
9
years. Further-
more, astronomers use as their unit of measurement a light year, which is
5,865,696,000,000 miles. We can now write this number as 5, 865, 696 10
6
miles.
The above reasoning is suﬃciently general to show that if N is any nonzero
whole number and k is any whole number, then
N 10
k
= the whole number obtained from N by attaching k
zeros to the right of the last digit of N.
5
The discerning reader would have noticed that the above derivation made implicit use
of the distributive law, which will be discussed presently in the next section.
1 Place Value 19
Finally, we deal with the issue of why the number 830159 is the same
number as 0830159, and is the same number as 000830159, etc. This is very
easy to explain because by the expanded form of a number,
000830159 = 0 10
8
+ 0 10
7
+ 0 10
6
+ 8 10
5
+
3 10
4
+ 1 10
2
+ 5 10
1
+ 9 10
0
= 830159 .
This observation is conceptually important in the understanding of the var-
ious algorithms in ¸3.
Exercise 1.1 Imagine you have to explain to a fourth grader that 43
100 = 4300. How would you do it?
Exercise 1.2 Imagine you have to explain to a ﬁfth grader why 48
500, 000, 000 = 24, 000, 000, 000. How would you do it?
Exercise 1.3 What number should be added to 946,722 to get 986,722?
What number should be added to 68,214,953 to get 88,214,953?
Exercise 1.4 What number should be added to 58 10
4
to get 63 10
4
?
What number should be taken from 52 10
5
to get 48 10
5
?
Exercise 1.5 Which is bigger? 4873 or 12001? 410
5
or 310
6
? 810
32
or 2 10
33
? 4289 10
7
or 10
11
? 765,019,833 or 764,927,919? Explain.
Exercise 1.6 Write each of the numbers 6100925730, 2300000000, and
7420000659 in expanded form.
Exercise 1.7 Show that for any nonzero whole number k, 10
k
> m10
k−1
for any single-digit number m.
Exercise 1.8 The following is the introduction to the concept of multi-
plication taken from a third grade textbook. On the side of the page is the
Vocabulary of the Day:
multiplication an operation using at least two numbers to
ﬁnd another number, called a product.
product the answer in multiplication.
Then in the text proper, one ﬁnds:
How many are in 4 groups of 6?
You can use multiplication to solve the problem.
Use cubes to model the problem and record the answer to the
problem:
2 The Basic Laws of Operations 20
Number of groups [ Number in each group [ Product
6 4 = 24
Further down:
If Helena practices singing 3 hours each day for a week, how
many hours will she practice altogether?
Find: 3 7
There is more than one way!
Method 1: You can use repeated edition to solve the problem.
3 + 3 + 3 + 3 + 3 + 3 + 3 = 21
Method 2: When the groups are equal, you can also write a
multiplication sentence.
7 3 = 21.
Write down your reaction to the appropriateness of such an introduction,
and compare your view with those of others’ in your class.
2 The Basic Laws of Operations
This section discusses the basic laws which govern the arithmetic operations
on whole numbers, e.g., why 23 + 79 = 79 + 23, or why (47 4) 5 =
47(45). The key point of such a discussion is always that two collections of
numbers which look superﬁcially diﬀerent are in fact equal, and this equality
is indicated by the ubiquitous equal sign “=”. Because the equal sign is one
of the sources of confusion in elementary school, let us ﬁrst deal with this
symbol.
The most important thing to remember is that while the meaning of
the equal sign does get more sophisticated as the mathematics gets more
advanced, there is no reason for us to learn everything about this symbol
all at once! We are starting from ground zero, the whole numbers, so the
meaning of “=” is both simple and unequivocal: two numbers a and b are
said to be equal if we can verify by counting that they are the same number.
2 The Basic Laws of Operations 21
For example, 4 + 5 = 2 + 7 because we count 4 objects and then 5 more
and get 9, whereas we count 2 objects and then 7 more and also get 9, so
this is what 4 + 5 = 2 + 7 means. So be sure to explain to your students —
again and again if necessary — that the equal sign between two collection
of whole numbers does not signify “do an operation to get an answer”. It
merely means:
check the numbers on both sides of the equal sign by counting to
verify that both sides yield the same number.
When we deal with fractions (Chapter 2) and decimals (Chapter 4), then
obviously we can no longer just count. The equality of two fractions or two
decimals will have to be more carefully explained (see ¸1 of Chapter 2 and
¸¸1 and 3 of Chapter 4).
Now we come to the main concerns of this section: the associative laws
and commutative laws of addition and multiplication, and the distributive
law. These are without doubt among the most hackneyed items you have
ever come across in mathematics. Your textbooks mention them with a
sense of noblesse oblig´e and can’t wait to get it over with, presumably they
believe you deserve better. Yet we are going to spend the next twenty
pages discussing exactly these laws without any apology. You are entitled
to know why. There are at least three reasons. The ﬁrst is that they are
used everywhere, sometimes implicitly without your being aware of them.
To bring home this point, let me just cite two of many instances where
we have used them in ¸1 in an “underhanded” manner: the proof that
2, 133, 070 100, 000 = 213, 307, 000, 000 implicitly made use of the distribu-
tive law, and the proof of (4) (that 10
m
10
n
= 10
m+n
) made implicit use of
the associative law of multiplication, as we shall explain below (see the dis-
cussion above (10)). Because these laws inﬁltrate every aspect of arithemtic
operations, your awareness of their presence would help you avoid making
incorrect pronouncements. Take, for instance, the simple problem of addi-
tion: 12 + 25 + 18. A quick way to compute this sum is to add 12 and 18
to get 30, and then add 30 to 25 to get 55. Now, some teachers probably
explain this shortcut to their second graders by saying
you just “ﬂy over” the middle number and add the two at both
ends,
but this would be incorrect because by convention,
2 The Basic Laws of Operations 22
in any expression involving arithemtic operations such as 12 +
25 + 18 or 4 17 25, it is understood that one only adds or
multiplies two neighboring numbers at a time: e.g., 12 + 25, or
25 + 18, or 4 17, or 17 25.
To add or multiply two number by “ﬂying over” another number is not a
permissible move in mathematics. But the possibility of computing 12+25+
18 as (12+18)+25 = 30+25 = 55 or 41725 as 17(425) = 17100 =
1700 will turn out to be entirely correct, because the apparent “ﬂying over”
can be precisely justiﬁed by the commutative and associative laws of addition,
and the commutative and associative laws of multiplication, respectively. As
a teacher, you have to be ready with the correct mathematical explanations
for such phenomena when the occasion calls for it.
A second reason is that knowing these basic laws can be helpful. Consider
the following simple problems:
(a) (87169 5) 2 = ?
(b) 10
7
6572 = ?
(c) 4 ([25 18] + [7 125]) = ?
These computations can be tedious if taken literally. For example, the deﬁ-
nition in (2) of ¸1 implies that (b) must be computed by adding 6572 to itself
10, 000, 000 times. Yet with a judicious application of the basic laws, these
computations can all be done eﬀortlessly in one’s head so that the answers
are, respectively, 871690, 65,720,000,000, and 5300. See the later discussions
in this section for the explanations.
A ﬁnal reason, and the most substantive one for our purpose, is that these
laws play a central role in this monograph. They weave in and out of the
ﬁve chapters, but are especially prominent in ¸¸3.3–3.4 of Chapter 1, ¸7.2 of
Chapter 2, and ¸3 of Chapter 5. For this reason, we take the opportunity to
clarify these foundational matters once and for all.
For addition, the two basic laws are the associative law and the commu-
tative law. These state that, if L, M and N are whole numbers,
(L + M) + N = L + (M + N) (5)
M + N = N + M (6)
These are nothing more than summaries of our collective experiences with
2 The Basic Laws of Operations 23
whole numbers, and we take their truth as an article of faith.
6
For
illustrations of some of these experiences, think of the addition of two num-
bers as combining two groups of discrete objects. Then the following pictures
describe the validity of the associative law (5) for L = 2, M = 4 and N = 7:
2

4

7

2

4

7

It is obvious from the pictures that whether we combine 2 and 4 ﬁrst and
then combine the sum with 7, or combine 2 with the sum of 4 and 7, we
get 13. Needless to say, the pictorial evidence for the truth of (5) can be
similarly obtained for other numbers. As to the commutative law (6), the
following pictures show that whether we combine 3 objects with 5 objects or
5 objects with 3 objects, either way we get 8 objects:
3

5

5

3
Again, the pictorial evidence is not dependent on the particular choice of
M = 3 and N = 5 and would hold equally well for other numbers.
We pause to comment on the equal sign in the context of the associative
law (5) to reinforce the deﬁnition of this symbol. What (5) asserts is that the
two numbers (L + M) + N and L + (M + N) are the same. In other words,
we count the left side by ﬁrst counting the group consisting of L objects and
M objects, and then continue the counting to include the next N objects; we
get one number. For the right side, we begin with L objects, then continue
the counting to include the group consisting of M objects and N objects,
thereby getting a second number. What (5) says is that these two numbers
are the same number.
6
In the proper axiomatic setup, both of (5) and (6) can be proved as theorems.
2 The Basic Laws of Operations 24
One consequence of the associative law of addition is that it clears up the
meaning of such common expressions as 4 + 3 + 7. This triple addition is a
priori ambiguous, because it could mean either (4 +3) +7 or 4 +(3 +7) .
(Note again that adding 4 to 7 before adding the result to 3 is not an option
because, by convention, only neighboring numbers are added.) But (5) tells
us that there is in fact no ambiguity because the two ways of adding are the
same. In general then, (5) leads to the conclusion that we don’t need to use
parentheses in writing the sum of any three whole numbers l +m+n , e.g.,
4 + 3 + 7 . Once this is noted, however, it should not be surprising that
we can draw the same conclusion about the sum of any four whole numbers
l + m + n + p. For example, let us show:
((l + m) + n) + p = l + ((m + n) + p),
where
the convention regarding parentheses is to do the innermost
parentheses ﬁrst and then systematically work one’s way out. Thus
((l + m) + n) + p means: add l to m, then add the result to n,
and ﬁnally add ((l + m) + n) to p.
Let us explain the preceding equality of four numbers by using (5) repeatedly.
Letting L = (l + m), M = n, and N = p in (5), we obtain
((l + m) + n) + p = (l + m) + (n + p). ()
Now let L = l, M = m, and N = (n + p) in (5) again, then the right side of
() becomes
(l + m) + (n + p) = l + (m + (n + p)).
Finally, letting L = m, M = n, and N = p and reading (5) from right to left,
we obtain m+(n +p) = (m+n) +p. Substituting this value of m+(n +p)
into the right side of the preceding equation, we obtain:
(l + m) + (n + p)) = l + ((m + n) + p). ()
Putting () and () together, we get
((l + m) + n) + p = l + ((m + n) + p),
as desired. Consequently, the meaning of l + m + n + p is also unambiguous
without the use of parentheses.
2 The Basic Laws of Operations 25
Because this kind of abstract, formal reasoning looks so facile and believ-
able, there is the danger that you would take it for granted. Let me therefore
make sure you are aware that there is substance beneath the smooth surface
of formalism. For example, if we let l = 5, m = 2, n = 8 and p = 1 and
compute both ((5 +2) +8) +1 and 5 +((2 +8) +1) directly to compare the
results step-by-step, it is far from obvious why they turn out to be equal at
the end:
((5 + 2) + 8) + 1 = (7 + 8) + 1 = 15 + 1 = 16
5 + ((2 + 8) + 1) = 5 + (10 + 11) = 5 + 11 = 16
This is the ﬁrst instance that we come across the paradoxical situation that,
by resorting to formal abstract reasoning, an argument for the general case
(which should be more diﬃcult) turns out to be more understandable than
the special case. I want to assure you that this will not be the last instance
where this happens.
The same reasoning shows that
given any collection of numbers, say 26 of them ¦a, b, c, , y, z¦,
we can unambiguously write
a + b + c + + y + z
without the use of any parentheses and, regardless of the order of
the addition of the numbers,
7
the result will be the same.
You cannot fail to notice at this point that in the very ﬁrst sum of ¸1, i.e.,
20000 + 500 + 40 + 1, we have already made implicit use of this fact. The
same can be said of the later expression (in ¸1) that
31536000 = 30000000 + 1000000 + 500000 + 30000 + 6000.
Needless to say, there are many other examples of using the associative law
of addition without mentioning it.
Essentially the same comments apply to the commutative law (6). For
example, if we have three numbers l, m and n, then all six expressions
l + m + n l + n + m m + l + n
m + n + l n + l + m n + m + l
7
Don’t forget the convention that one can only add two neighboring numbers at a time.
2 The Basic Laws of Operations 26
are the same. (Notice that we have already made use of the fact that the
use of parentheses is not needed for the addition of three numbers !) Let us
show, for instance, that l + m + n = n + m + l by applying (6) repeatedly
to two numbers at a time:
l + m + n = m + l + n = m + n + l = n + m + l
Of course the same is true of the addition of any collection of whole numbers:
the order of appearance of the whole numbers in any ﬁnite sum is
unimportant.
By now, you undoubtedly appreciate one aspect of the laws (5) and (6):
they are tremendous labor saving devices. If we have ﬁve numbers 2, 4, 5 11,
3, then there are 120 ways of adding them (try it!). But we only have to do
one of them, say, ((4 + 5) + 11) + (2 + 3) = (9 + 1) + 5 = 20 + 5 = 25, and
there is no need to do any of the other 119 sums because they would all be
equal to 25. This then justiﬁes the writing of 2 + 5 + 4 + 11 + 3 = 25.
Before proceeding further, it may be time to tie up the loose end left open
earlier as to why it is permissible to ﬁrst add 12+18 in the sum 12+25+18.
In detail, we are arguing that
12 + 25 + 18 = (12 + 18) + 25.
Observe ﬁrst of all that the associative law of addition is already used in
writing 12+25+18 without the use of parentheses; we may insert parentheses
any which way we wish. So with this understood, we give the proof:
12 + 25 + 18 = 12 + (25 + 18)
= 12 + (18 + 25) (commutative law)
= (12 + 18) + 25 (associative law)
Exactly as claimed.
We should emphasize that argument of this sort is primarily for your ben-
eﬁt as a teacher and should not be taken as a statement that good teaching
means always explaining details of this kind. What it does say is that, as
a teacher, you should be ready with an explanation. In the fourth or ﬁfth
grade, for instance, there will be occasions for you to introduce this kind of
2 The Basic Laws of Operations 27
reasoning to your students, gently. But a teacher must exercise good judg-
ment in not overdoing anything.
Next we turn to the multiplication of whole numbers as deﬁned in (2) of
¸1. We have two similar laws, the associative law of multiplication and the
commutative law of multiplication: for for any whole numbers L, M and N,
(LM)N = L(MN) (7)
MN = NM (8)
where we have made use of the
notational convention: when letters are used to stand for num-
bers, the multiplication sign is omitted so that “MN” stands for
“M N”.
Before discussing the empirical evidence behind (7) and (8), we can
demonstrate their power by applying them to problems (a) and (b) posed at
the beginning of this section. First, (871695)2 is equal to 87169(52),
by the associative law (7). Since 52 is 10, the triple product is immediately
seen to be equal to 871690. This disposes of (a). Next, 10
7
6572 = 657210
7
by the commutative law (8), but the right side is 65,720,000,000 by the con-
clusion we arrived at near the end of ¸1, to the eﬀect that
N 10
k
= the whole number obtained from N by attaching k
zeros to the right of the last digit of N.
This then disposes of (b).
The way we have just dealt with (a) and (b) aﬀords an excellent op-
portunity to remind you of the importance of using deﬁnitions exactly as
given. It would have been very easy, for example, for you to equate in your
minds without thinking that 10
7
6572 = 6572 10
7
and then conclude
that the result is 65,720,000,000. What we are doing here is to intentionally
bring the underlying reason (the commutative law (8)) for the validity of
10
7
6572 = 657210
7
to your attention: by deﬁnition, 10
7
6572 is adding
6572 to itself 10,000,000 times, whereas 6572 10
7
is adding 10,000,000 to
itself 6572 times. They look at ﬁrst glance, at least, to be quite diﬀerent
animals. The equality of these two products may not seem important to you
because you have taken it for granted. But if you do not begin to take note
2 The Basic Laws of Operations 28
of the role of (8) in arithmetic, you would not be able to follow the reasoning
in later discussions when the direction of the discussion is determined by (8),
e.g., in ¸3.4 when we show that the two interpretations of division yield the
same result.
As with the corresponding laws for addition, one may regard (7) and (8)
as summaries of empirical experiences and accept them on faith. Here is the
kind of pictorial evidence that is easily available:
Activity: Verify by direct counting that (8) is valid for M
and N between 1 and 5 inclusive (in symbols: 1 ≤ M, N ≤ 5),
by using rectangular arrays of dots to represent multiplication of
whole numbers as in the discussion below (2) in ¸1.
Activity: Verify by direct counting that (7) is valid for L, M,
N between 1 and 5 inclusive (in symbols: 1 ≤ L, M, N ≤ 5), by
using 3-dimensional rectangular arrays of dots to represent triple
products of whole numbers.
For later needs in this monograph (cf. the discussions of fractions and
decimals in Chapters 2 and 4), the representation of multiplication by dots
would be inadequate and it would be more suitable to use a area model.
For example, 3 5 is represented as a rectangle with “vertical” length 3
(corresponding to the ﬁrst number) and “horizontal” length 5 (corresponding
to the second number):
(9)
According to this description of the area model, the product 5 3 would be
represented by the area of a rectangle with “vertical” length 5 and “horizon-
tal” length 3:
2 The Basic Laws of Operations 29
Because these two rectangles can be obtained from each other by a 90
◦
ro-
tation, they have the same area. This then give a pictorial “explanation” of
3 5 = 5 3. As usual, this discussion is independent of the choice of the
numbers 3 and 5 and would be the same for any two numbers m and n.
In ¸4, we will present yet another interpretation of the multiplication of
whole numbers that will be important when we come to fractions. We cannot
give this interpretation here because it involves a deeper understanding of
what a “number” is.
A word about “area” would be appropriate: A detailed discussion of this
concept will not be attempted here, nor will it be needed, except to mention
that, by convention, we agree to let the area of the unit square (the square
with each side equal to 1) to be just 1, so that the area of the rectangle in
(9) is 15 because precisely 15 unit squares tile (or pave) the rectangle.
It is worthwhile to point out that the area model of multiplication pro-
vides the mathematical underpinning of the manipulative Base Ten Blocks.
Most likely you have used this manipulative in your classroom to facilitate
the learning of multiplication. We would like to add a passing comment that
while Base Ten Blocks, like other manipulatives, can be helpful, students
should not be allowed to become dependent on it. The real challenge in
a mathematics classroom is still to learn the mathematics, not manipulate
manipulatives.
2 The Basic Laws of Operations 30
The product of three whole numbers l, m and n can be represented as
the volume of a rectangular solid. For example, (3 5) 2 is the volume of
the rectangular solid of height 2 built on the rectangle in (9) above:
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
¨
¨
¨
¨
¨¨
Similarly, 3 (5 2) = (5 2) 3 by the commmutative law (8), so that
3 (5 2) is the volume of the solid of height 3 built on the rectangle with
“verticle” length 5 and “horizontal” length 2:
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
¨
The equality of the volumes of these two solids — because one is obtained
from the other by a rotation in space — is then the pictorial evidence for the
truth of (3 5) 2 = 3 (5 2). We remark as in the case of area that
we shall not go into the precise deﬁnition of volume but will only use it in
an intuitive way. Again, the concrete numbers 3, 5, and 2 of the preceding
argument can be replaced by any triple of numbers in (7).
Of course, as in the case of addition, there are more general versions of
(7) and (8) for arbitrary collections of numbers. For example, the previous
discussion of the associative law and commutative law for addition ((5) and
(6)) applies to multiplication verbatim, and we know that
2 The Basic Laws of Operations 31
the product of any collection of numbers can be written unam-
biguously without the use of parentheses and without regard to
order.
Thus, for any four numbers, say l, m, n, p, their product can be written in
any of the following 24 ways:
lmnp lmpn lnmp lnpm lpmn lpnm
mlnp mlpn mnlp mnpl mpln mpnl
nlmp nlpm nmlp nmpl nplm npml
plmn plnm pmln pmnl pnlm pnml
and all of them are equal to ((lm)n)p. To drive home the point that ﬁrst
surfaced in connection with the associative law of addition, let us use four
explicit numbers — say l = 7, m = 3, n = 2 and p = 4 — to illustrate the
nontrivial nature of, for example, mlpn = plnm:
(7 3) (2 4) = 21 (2 4) = 21 8 = 168
(4 (7 2)) 3 = (4 14) 3 = 56 3 = 168 .
At the risk of harping on the obvious, note that none of the intermediate
steps of the two computations look remotely similar, and yet miraculously
the ﬁnal results are identical.
Now we are in a position to point out in what way the proof of (4) in ¸1,
namely, 10
m
10
n
= 10
m+n
, implicitly made use of the associative law of
multiplication. Properly speaking, its proof should go as follows: Because we
may write a product of m numbers without the use of parentheses, we have
10
m
= 10 10 10

m
. Similarly, 10
n
= 10 10 10

n
. Therefore,
10
m
10
n
= (10 10 10

m
) (10 10 10

n
)
However, we also know that a product of m + n numbers can be written
without the use of parentheses. Consequently,
10
m
10
n
= 10 10 10

m+n
,
and the right side is just 10
m+n
. This shows 10
m
10
n
= 10
m+n
.
2 The Basic Laws of Operations 32
Finally, the distributive law connects addition with multiplication. It
states that for any whole numbers M, N, and L,
M(N + L) = MN + ML (10)
Recall in this connection
the convention that in the expression MN + ML, we multiply
the numbers MN and ML ﬁrst before adding.
Again, like the other laws we have discussed so far, the distributive law
(10) is nothing more than a summary of common sense. Pictorially, we have:
Activity: Directly verify (10) using rectangular arrays of dots
to represent multiplication for 1 ≤ L, M, N ≤ 5.
One can also use the area model: if M = 3, N = 2 and L = 4, then 3(2 +4)
is the area of the following big rectangle:
On the other hand, 3 2 is the area of the left rectangle and 3 4 is the
area of the right rectangle. Thus 3(2 + 4) = (3 2) + (3 4). Again, the
essence of this picture is unchanged when 2, 3, and 4 are replaced by other
triples of numbers.
Clearly, the distributive law generalizes to more than three numbers. For
example:
m(a + b + c + d) = ma + mb + mc + md
for any whole numbers m, a, b, c and d. This can be seen by applying the
distributive law (10) twice, as follows:
m(a + b + c + d) = m(¦a + b¦ +¦c + d¦)
= m¦a + b¦ + m¦c + d¦
= ¦ma + mb¦ +¦mc + md¦
= ma + mb + mc + md
2 The Basic Laws of Operations 33
Observe that each step in the preceding calculation depends on the earlier
discussion of the legitimacy of writing a + b + c + d without parentheses.
As mentioned earlier, the distributive law is the glue that binds addition
+ and multiplication together. Despite its obvious importance, it seems
to be the least understood among the three laws and for many, a ﬁrm grasp
of this law proves elusive. We urge you to spare no eﬀort in learning to use
it eﬀectively. Spending more time with some of the exercises at the end of
this section may be the answer.
A common mistake in connection with the distributive law is to remember
(10) only in the form of “the left side of (10) is equal to the right side” (i.e.,
M(N + L) = MN + ML) without realizing that (10) also says “the right
side of (10) is equal to the left side” (i.e., MN + ML = M(N + L)). In
other words, given 35 (72 + 29), most people recognize that it is equal
to (35 72) + (35 29), but when (35 72) + (35 29) is given instead,
they fail to see that it is equal to 35 (72 +29). In practice, the latter skill
may be the more critical of the two, and there is an mathematical reason
for this. In terms of the preceding example, (35 72) + (29 35) involves
two multiplications ( 35 72 and 29 35 ) and one addition, whereas
35 (72 + 29) involves only one multiplication and one addition. Because
multiplication is in general more complicated, it is preferable to multiply as
little as possible and therefore preferable to compute 35 (72 +29) rather
than (35 72) +(29 35). Therefore, to recognize (35 72) +(29 35) =
35 (72 + 29) is to be able to achieve a simpliﬁcation.
To further pin down the last idea, we bring closure to this section by
doing problem (c) posed near the beginning: 4 (25 18 +7 125) = ? We
know 125 = 5 25, so that 7 125 = 7 5 25 = 25 [7 5]. Therefore:
4 (((25 18 + 7 125) = 4 (25 18 + 25 [7 5])))
= 4 25 (18 + [7 5]) (distributive law)
= 100 53 = 53 100
= 5300.
It should be clear that the whole computation can be done by mental math.
On the other hand, we would be looking at calculating three multiplications
if we don’t appeal to the distributive law: 25 18, 7 125, and 4 1325,
where 1325 = 25 18 + 7 125.
2 The Basic Laws of Operations 34
Moral: Be sure you know MN + ML = M(N + L).
We began the discussion of order among whole numbers (i.e., which whole
number is bigger) in ¸1. We can now conclude that discussion. Recall that
given two whole numbers a and b, the inequality a < b is deﬁned to mean
that in the counting of the whole numbers starting with 0, 1, 2, . . . , the
number a precedes b. For the convenience in logical arguments, we wish
to express this deﬁnition diﬀerently:
The statement a < b is the same as the statement: there is a
nonzero whole number c so that a + c = b.
Before giving the reason for this assertion, we explain what is meant by the
two conditions being the same. This is a piece of mathematical terminology
that signiﬁes that both of the following statements are true:
If a < b, then there is a nonzero whole number c so that
a + c = b.
Conversely, if there is a nonzero whole number c so that a+c = b,
then a < b.
In concrete situations, both statments are quite obvious. For example,
7 < 12 means that in counting from 7, we have to go 5 more steps before
we get to 12, so 7 + 5 = 12. Conversely, if we know 7 + 5 = 12, then we
must go 5 more steps from 7 before we get to 12, so 7 < 12. The general
reasoning is not much diﬀerent. Given a < b, we know by deﬁnition that
a precedes b, so that in the counting of the whole numbers, after we get
to a, we need to go (let us say) c more steps before getting to b, and c is
not zero. This implies a + c = b. Conversely, suppose a + c = b is given,
with c > 0, then after counting a objects, we have to count c more objects
before we get b objects. So by the deﬁnition of “smaller than”, we know a < b.
At this juncture, it is time to introduce the number line in order to give
a geometric interpretation of inequalities. Fix a straight line and designate
a point on it as 0. To the right of 0, mark oﬀ equally spaced points and call
them 1, 2, 3, 4, etc., as on a ruler. Thus the whole numbers are now identiﬁed
with these equally spaced points on the right side of the line. It is convenient
to single these points out by markers (notches), again as on a ruler, and to
identify the points with the markers. We explicitly call attention to the fact
2 The Basic Laws of Operations 35
that the counting of the whole numbers (as was done in ¸1) corresponds to
the progression of the markers to the right of the line starting with the initial
marker 0, as shown below. (Until Chapter 5, we will have no need for the
part of the line to the left of 0):
0 1 2 3 4 5 etc.
It follows from the way the number line is drawn that
two whole number a and b satisfy a < b precisely when the
position of a on the number line is to the left of that of b.
Here is a pictorial representation of the situation:
a b
a < b
A further geometric interpretation can be given: given a < b as shown,
suppose a + c = b, then c is precisely the number of markers between a and
b. This follows immediately from the way the whole numbers are positioned
on the number line.
Activity: Verify the last statement about c for some concrete
numbers such as a = 8, b = 21, or a = 86, b = 95.
The following facts about inequalities are well-known:
a + b < a + c is the same as b < c
a ,= 0, ab < ac is the same as b < c
a < b and c < d implies a + c < b + d
a < b and c < d implies ac < bd
(11)
We will only give the explanation for the second assertion (the most
diﬃcult of the four) and leave the rest as exercises.
Activity: Convince yourself that 2a < 2b is the same as a < b,
and that 3a < 3b is the same as a < b. Do it both numerically as
well as on the number line.
2 The Basic Laws of Operations 36
Now the proof of the second assertion. Given a ,= 0. First we prove that
ab < ac implies b < c.
There are three possibilities between the whole numbers b and c : (A) b = c,
(B) c < b, and (C) b < c. We know ahead of time that only one of the three
possibilities holds. In order to show that (C) is the correct conclusion, all
we need to do is to show that both (A) and (B) are impossible. Now if (A)
holds, then ab = ac, which is contrary to the assumption that ab < ac. We
have therefore eliminated (A). If (B) holds, then c < b, and there will be a
nonzero whole number l so that c +l = b. It follows that a(c +l) = ab, which
implies ac + al = ab, by the distributive law. But both a and l are nonzero,
so al > 0 and therefore ac < ab. This is also contrary to the assumption that
ab < ac. Hence (B) is also eliminated. It follows that (C) is the only possible
conclusion.
Next, we prove the converse, i.e.,
b < c implies ab < ac.
Now b < c implies that b + l = c for some nonzero l. Thus a(b + l) = ac, so
that ab + al = ac again by the distributive law. Because al is nonzero, the
last equation means ab < ac, as desired. This completes the proof.
Exercise 2.1 Elaine has 11 jars in each of which she put 16 ping pong
balls. One day she decided to redistribute all her ping pong balls equally
among 16 jars instead. How many balls are in each jar? Explain.
Exercise 2.2 Before you get too comfortable with the idea that everything
in this world has to be commutative, consider the following. (i) Let A
1
stand
for “put socks on” and A
2
for “put shoes on”, and let A
1
◦ A
2
be “do A
1
ﬁrst, and then A
2
”, and similarly let A
2
◦ A
1
be “do A
2
ﬁrst, and then A
1
”.
Convince yourself that A
1
◦ A
2
does not have the same eﬀect as A
2
◦ A
1
. (ii)
For any whole number k, let B
1
(k) be the number obtained by adding 2 to
k, and B
2
(k) be the number obtained by multiplying k by 5. Show that no
matter what the number n may be, B
1
(B
2
(n)) ,= B
2
(B
1
(n)).
Exercise 2.3 Find shortcuts to do each of the following computations and
give reasons (associative law of addition? commutative law of multiplication?
etc.) for each step: (i) 833 + (5167 + 8499), (ii) (54 + 69978) + 46, (iii)
(25 7687) 80, (iv) (58679 762) +(58679 238), (v) (4 4 4 4)
(5 5 5 5 5), and (vi) 64 125, (vii) (69 127) + (873 69), M
(viii) (((([125 24] 674) + ([24 125] 326))).
2 The Basic Laws of Operations 37
The purpose of the last exercise is not to get you obsessed with tricks in
computations everywhere. Tricks are nice to have, but they are not the main
goal of a mathematics education, contrary to what some people would have
you believe. What this exercise tries to do is, rather, to make you realize that
the basic laws of operation discussed in this section are more than empty,
abstract gestures. They have practical applications too.
Exercise 2.4 Prove the remaining three assertions in (11).
Exercise 2.5 Prove that both assertions in (11) remain true if the strict
inequality symbol “<” is replced by the weak inequality symbol “≤”.
Exercise 2.6 Let m and n be a 3 digit number and a 2 digit number,
respectively. Can mn be a 4 digit number? 5 digit number? 6 digit number?
7 digit number?
Exercise 2.7 Let m and n be a k digit number and an digit number,
respectively, where k and are nonzero whole numbers. How many digits
can the number mn be? List all the possibilities and explain.
Exercise 2.8 Suppose you have a calculator which displays only 8 digits
(and if you have a fancy calculator, you will be allowed to use only 8 digits!),
but you have to calculate 856164298 65. Discuss an eﬃcient method to
make use of the calculator to help with the comutation. Explain. Do the
same for 376241048 872.
Exercise 2.9 Let x and y be two whole numbers. (i) Explain why (x +
y)(x + y) = x(x + y) + y(x + y). (ii) Explain why (x + y)(x + y) =
xx+xy +yx+yy. (iii) Explain why (x+y)(x+y) = x
2
+2xy +y
2
, where
as usual x
2
means xx and y
2
means yy.
Exercise 2.10 Let x and y be two whole numbers. Explain why (x −
y)(x + y) = (x −y)x + (x −y)y.
Exercise 2.11 The following is how a fourth grade textbook introduces
the associative law of multiplication.
Ramon buys yo-yos from two companies, He buys six different
styles from each company and gets each style in 4 different
colors. How many yo-yos does he buy in all?
Find 264 to solve. You can use the associative property
to multiply three factors. The grouping of the numbers does not
affect the answer.
Step 1: Use parentheses to show grouping.
2 6 4 = (2 6) 4
3 The Standard Algorithms 38
Step 2: Look for a known fact to multiply.
2 4 is a known fact.
Step 3: Use the Commutative Property to change the order, if
necessary.
(2 6) 4 = (6 2) 4
= 6 (2 4)
= 6 8 = 48
Write down your reaction to such an introduction, and compare with
those of others’ in your class.
3 The Standard Algorithms
By an algorithm, we mean an explicitly deﬁned, step-by-step computational
procedure which has only a ﬁnite number of steps. The purpose of this
section is to describe as well as provide a complete explanation of the so-
called standard algorithms for the four arithmetic operations among whole
numbers.
At the outset, we should make clear that there is no such thing as the
unique standard algorithm for any of the four operations +, −, , ÷, be-
cause minor variations in each step of these algorithms not only are possible,
but have been incorporated into the algorithms by various countries and even
diﬀerent ethnic groups. The underlying mathematical ideas are however al-
ways the same, and it is these ideas that are the focus of our attention here.
For this reason, the nomenclature of “standard algorithms” is eminently jus-
tiﬁed. This is not to say that the algorithms themselves — the mechanical
procedures — are of no interest. On the contrary, they are, because compu-
tational techniques are an integral part of mathematics. Furthermore, the
conciseness of these algorithms, especially the multiplication algorithm and
the long division algorithm, is a marvel of human invention. One of the goals
of this section is to make sure that you come away with a renewed respect for
them. Nevertheless, we shall concentrate on the mathematical ideas behind
them as you are likely to be less familiar with these ideas.
3 The Standard Algorithms 39
A fundamental question about these arithmetic algorithm is why you
should bother to learn them. Take a simple example: what is 17 12 ?
By deﬁnition, this is 12 added to itself 17 times and one school of thought
would have you count 17 piles of birdseed with 12 in each pile. But what
about 34, 609 549, 728 ? Because brutal counting is less than attractive
for numbers of this magnitude, a shortcut is clearly called for. This is where
algorithms come in: they provide a shortcut in lieu of direct counting. There-
fore at the outset, the eﬃciency of an algorithm — how to get the answer
as simply and quickly as possible — is an overriding concern. But one could
push this argument further. Why worry about eﬃciency if pushing buttons
on a calculator may be the most eﬃcient way to make a computation such as
34, 609 549, 728 ? There are at least two reasons why a strict reliance on
the calculator is inadequate. First, without a ﬁrm grasp of the place value of
our numeral system and the mathematical underpinning of the algorithms, it
would be impossible to detect mistakes caused by pushing the wrong buttons
on the calculators.
8
A more important reason is that learning the reasoning
behind these eﬃcient algorithms is a very compelling way to acquire many
of the fundamental skills in mathematics, including abstract reasoning and
symbolic manipulative skills. These are skills absolutely essential for the
understanding of fractions and decimals in the subsequent chapters. Much
more is true. The present crisis in the learning of algebra in schools would
have been largely eliminated if students were properly taught the logical rea-
soning behind the algorithms in the upper elementary and middle schools.
9
Please keep all this in mind when you learn the mathematics surrounding
the algorithms and especially when you teach them to your students.
There is a kind of leitmotif all through the algorithms, which can be
roughly described as follows:
To perform a computation with an n digit number, break the com-
putation into n steps so that each step involves only one digit of
the given number.
The precise meaning of this statement will be made clear with each algorithm,
but the overall idea is that those simpler single-digit computations can all be
8
I trust that it would be unnecessary to recount the many horror stories related to
ﬁnger-on-the-wrong-button.
9
This conclusion is based on the fact that, in mathematics, learning does not take place
without a solid foundation.
3 The Standard Algorithms 40
carried out routinely without thinking. The last sentence calls for some com-
ments as it runs counter to some education theories which a segment of the
education community holds dear. The fact that a crucial part of mathematics
rests on the ability to break down whole concepts into discrete sub-concepts
and sub-skills must be accepted if one hopes to learn mathematics. This is
the very nature of mathematics, and no amount of philosophical discussion
would change that. A second point concerns the uneasiness with which some
educators eyes the routine and nonthinking nature of an algorithm. It is the
very routineness that accounts for the fact that these algorithms get used; it
guarantees an easy way to get results. If we teach these algorithms without
emphasizing their routine character, we would be falsifying mathematics.
As to the non-thinking aspect of these algorithms, there is at present
a perception that if anything can be done without thinking, then it does
not belong in a mathematics classroom. This is wrong. If mathemati-
cians are forced to do mathematics by having to think every step of the way,
then little mathematics of value would ever get done and all research math-
ematics departments would have to close shop. What is closer to the truth
is that deep understanding of a topic tends to reduce many of its sophis-
ticated processes to simple mechanical procedures. The ease of executing
these mechanical procedures then frees up mental energy to make possible
the conquest of new topics through imagination and mathematical reasoning.
In turn, much of these new topics will (eventually) be once again reduced
to routine or nearly routine procedures, and the process then repeats itself.
There is nothing to fear about the ability to execute a correct mathematical
procedure with ease, i.e., without thinking. More to the point, having such
an ability in the most common mathematical situations is not only a virtue
but an absolute necessity. What one must fear is limiting one’s mastery of
such procedures to only the mechanical aspect and ignoring the mathemat-
ical understanding of why the procedures are correct. A teacher’s charge in
the classroom is to promote both facility with procedures and the ability to
reason. In the teaching of these algorithms, we should emphasize both their
routine nature as well as the logical reasoning that lies behind the procedures.
The preceding discussion is about the kind of mathematical understand-
ing teachers of mathematics must have in approaching the basic arithmetic
algorithms. The pedagogical issue of how to introduce these algorithms to
children in the early grades is something that lies outside the scope of this
monograph and needs to be treated separately. There is however a discus-
3 The Standard Algorithms 41
sion of this issue concerning the addition and multiplication algorithms in
the article, “Basic skills versus conceptual understanding”, in the Fall 1999
issue of the American Educator, p. 14; it is also accessible at
http://www.aft.org/american educator/fall99/wu.pdf
3.1 An addition algorithm
Given any two numbers, say 4 and 7, to ﬁnd the sum 4+7 is a simple matter
in principle: start with 4, we count 7 more times until we reach 11, and that
would be the sum. There is a graphic representation of this in terms of the
number line:
E
4
E
7
0 4
4 + 7
11
Similarly, here is a graphic representation of 9 + 5 = 14:
9
E E
5
9 + 5
0 5 9 14
For many purposes, it is more convenient to take a slightly diﬀerent point
of view. Denote the line segment from 0 to 7 by [0, 7]. It is natural to call 7
the length of [0, 7]. Simlarly, we deﬁne the length of the segment [0, n] from
0 to the whole number n to be n. We can now deﬁne the length of more
general line segments by treating the number line as an “inﬁnite ruler”, as
follows. For any line segment [x, y] from a point x to another point y on the
number line, — where it is understood from the notation [x, y] that y is to
the right of x — we say the length of [x, y] is n for a whole number n if,
by sliding [x, y] to the left along the number line until x rests at 0, the right
3 The Standard Algorithms 42
endpoint y rests at n. With the notion of length at our disposal, we can
describe another way to ﬁnd the sum of 4 +7: concatenate the two segments
[0, 4] and [0, 7] in the sense of placing them on a straight line end-to-end,
then the length of the concatenated segment is the sum of 4 + 7. Thus:

[0,4]

[0,7]
11
' E
Notice that concatenating the left endpoint of [0, 7] to the right endpoint of
[0, 4] corresponds exactly to “start with 4, we count 7 more times until we
reach 11”.
Similarly, we can concatenate [0, 9] and [0, 5] to get 9 + 5 = 14:

[0,9]

[0,5]
14
' E
This discussion continues to be meaningful when 4 + 7 and 9 + 5 are re-
placed by a + b for any a + b where a and b are whole numbers. In principle
then, addition is simple.
Now look at 4502 +263. It can be rather trying to count 263 times start-
ing from 4502 before getting an answer. (Try it!) However, a special feature
of the Hindu-Arabic numeral system, namely, the fact that its numerals “al-
ready come pre-packaged”, renders such a desperate act completely unneces-
sary. Let us use a simpler example of addition to explain what is meant by
the phrase in quotes. Suppose we have two sacks of potatoes, one containing
34 and the other 25, and we want to know how many potatoes there are
altogether. One way is to dump the content of both sacks to the ground and
start counting, which is what we called “the desperate act”. But suppose
upon opening the sacks, we ﬁnd that in each sack, the potatoes come in bags
3 The Standard Algorithms 43
of 10: the sack of 34 potatoes is put in 3 bags of 10 each plus 4 stray ones,
while the sack of 25 is put in 2 bags of 10 each plus 5 stray ones. Therefore
an intelligent way to count the total number of potatoes is to ﬁrst count the
total number of bags of 10’s (3 + 2 is 5, so there are 5 bags of 10 each), and
then count the stray ones separately (4 + 5 is 9, and so there are 9 extra).
Thus the total is 5 bags of 10 each, plus 9 extra ones, which puts the total at
59. This is exactly the idea behind the addition algorithm because
the number 3 in 34 — being in the tens place — signals that there are 3 tens,
and the 2 in 25 signals that there are 2 tens. Adding 2 and 3 to get 5, we
know that there are 5 tens in the total. Adding 4 to 5 then rounds oﬀ the
whole sum, and we get 34 + 25 = 59.
The standard addition algorithm is nothing more than a formal elabora-
tion of this simple idea. It says:
The sum of two whole numbers can be computed by lining them up
digit-by-digit, with their ones digits in the extreme right column,
and adding the digits column-wise, starting with the right column
and moving to the left; in case no digit appears in a certain spot,
assume that the digit is 0.
Schematically then for 4502 + 263:
4 5 0 2
+ 2 6 3
4 7 6 5
(12)
Thus starting from the ones digit (right column), we have: 2+3 = 5, 0+6 = 6,
5 + 2 = 7, and since there is no digit in the spot below the number 4, the
algorithm calls for putting a 0 there and the sum is 4+0 = 4 for that column.
Observe how the addition algorithm illustrates the leitmotif mentioned near
the beginning of the section: the addition 4205 + 263 is reduced to the
calculation of four single-digit additions: 4 + 0, 5 + 2, 0 + 6, and 2 + 3.
The following discussion of addition will be restricted to this particular
version of the algorithm. In due course, we shall discuss why the algorithm
moves from right to left, which some people consider unnatural.
First of all, when this algorithm is taught in the classroom, the main
emphasis is usually not on simple cases such as 4502 +263 where the sum of
3 The Standard Algorithms 44
the digits in each column remains a single-digit number, but on cases such
as 69 + 73 where the process of “carrying” to the next column takes place.
In this monograph, we reverse the emphasis by spending more time on the
simple case before dealing with the more complicated case. There is a good
reason for this decision: it is in the simple case that one gets to see with
unobstructed clarity the main line of the logical reasoning, and when that
is well understood, the more complicated case — which is nothing but the
simple case embellished with a particular technique — tends to follow easily.
To underscore the fact that each step of the algorithm is strictly limited to
the consideration of a single digit without regard to its place value, consider
865 + 32:
8 6 5
+ 3 2
8 9 7
(13)
Notice that the third column from the right of (12) and the rightmost column
of (13) are identical:
5
+ 2
7
(14)
Yet, in terms of place value, we know that in the context of (12), the addition
fact in (14) actually stands for
5 0 0
+ 2 0 0
7 0 0
because the 5 in 4502 stands for 500 and the 2 in 263 stands for 200. By
contrast, in the context of (13), the addition fact in (14) is literally true: it
is just 2 +5 = 7. As far as the algorithm is concerned, however, the addition
fact (14) is carried out in exactly the same way in (12) or (13), without regard
to this diﬀerence.
Would this digit-by-digit feature of the algorithm corrupt students’ un-
derstanding of place value? Not if students are made to understand that,
far from a defect, this digit-by-digit feature is a virtue for the purpose of
easy computation. Had the procedure of the algorithm treated each digit
diﬀerently according to its place value, the algorithm would lose much of its
simplicity: imagine that for the ones place you do one thing, for the tens
3 The Standard Algorithms 45
place you do another, and for the hundreds place you do yet another, etc.
How eﬃcient can the algorithm be in that case? Moreover, in learning the
procedural aspect of the algorithm, students should at the same time achieve
the understanding that the algorithm is correct precisely because of place
value considerations. To see this, recall:
865 = 8 10
2
+ 6 10
1
+ 5 10
0
32 = 0 10
2
+ 3 10
1
+ 2 10
0
(15)
(Notice that we have made use of the trivial fact 32 = 032 mentioned at the
end of ¸1.) Recall an earlier comment made also in ¸1 about the occasional
advantage of explicitly writing down all the powers of 10 in the expanded form
of a number. We will see how this more clumsy notation lends conceptual
clarity to what we have to do. We add these two equations. The left sides
add up to 865 + 32, of course. What about the right sides? We can add
“vertically”:
8 10
2
+ 0 10
2
= (8 + 0) 10
2
(= 8 10
2
)
6 10
1
+ 3 10
1
= (6 + 3) 10
1
(= 9 10
1
)
5 10
0
+ 2 10
0
= (5 + 2) 10
0
(= 7 10
0
) ,
where we have made use of the distributive law (10) three times in succession.
(It may be instructive for you to read the exhortation near the end of ¸2 about
the need to know that MN + ML = M(N + L). Because the left side of
the sum of equations equals the right side of this sum, we get
865 + 32 = (8 + 0) 10
2
+ (6 + 3) 10
1
+ (5 + 2) 10
0
(= 8 10
2
+ 9 10
1
+ 7 10
0
= 897 ) ,
(16)
which is seen to be a parallel description of the addition algorithm in (13).
We now see that
the addition algorithm is the method of adding two numbers by
adding the digits corresponding to the same power of 10 when the
two numbers are written out in their expanded forms.
In particular, we see why the algorithm calls for replacing the empty spot
under 8 with 0: it is none other than an abbreviated statement of 8 + 0 = 8
in (16). The reasoning is valid in general and is by no means restricted to
this special case of 865 + 32.
3 The Standard Algorithms 46
Activity: Give a similar explanation of (12).
The preceding explanation of the addition algorithm — by this we mean
the main ideas but not necessarily the precise notational formalism — would
be adequate in most classroom situations. It is important to realize, however,
that there are subtle gaps in the reasoning above, so that in the interest of
a complete understanding, we proceed to ﬁll in these gaps.
From (15), we have:
865 + 32 = ¦8 10
2
+ 6 10
1
+ 5 10
0
¦+
¦0 10
2
+ 3 10
1
+ 2 10
0
¦
As noted in connection with the associative and commutative laws of addition
(5) and (6), we can ignore the braces and rearrange the order of summation
of these six terms. Thus,
865 + 32 = ¦8 10
2
+ 0 10
2
¦ +¦6 10
1
+ 3 10
1
¦+
¦5 10
0
+ 2 10
0
¦
It is at this point that we can apply the distributive law (10) to conclude
(16). Thus what we said above about “adding vertically” in (15) is in fact an
application in disguise of the associative and commutative laws of addition.
Pedagogical Comments: Should one emphasize discussion of the
above type in an elementary school classroom? Few questions in education
can be answered with absolute certainty, but there are at least two reasons
why an elaborate discussion of the associative law and commutative law in,
say, K–5 might interfere with a good mathematics education. First, such
explanations tend to be somewhat tedious and students might lose interest
and, second, such details might obscure the main thrust of the argument
which is encapsulated in (16). As a teacher, however, you owe it to yourself
to understand this kind of details. This is because, on the one hand, intel-
lectual honesty demands it and, on the other, you must be prepared in case
a precocious youngster presses you for the complete explanation. End of
Pedagogical Comments.
For the case of 865 + 32, it is possible to give a naive explanation of the
addition algorithm in terms of money. Imagine that someone has
8 hundred-dollar bills,
6 ten-dollar bills, and
5 one-dollar bills,
thus $865 altogether. Later she acquires another stack of bills consisting of
3 The Standard Algorithms 47
3 ten-dollar bills, and
2 one-dollar bills,
thus another $32. To ﬁnd out how much money she has altogether, she
decides on the following strategy: collect all the hundred-dollar bills, then
collect all the ten-dollar bills, and ﬁnally collect all the one-dollar bills. She
ﬁnds that she now has
8 (= 8 + 0) hundred dollar bills
9 (= 6 + 3) ten-dollar bills
7 (= 5 + 2) one-dollar bills
So she has $897. Exactly as in (13).
Before proceeding further, the question must be raised as to why it is not
suﬃcient to understand the addition algorithm in terms of money alone, and
why we must go through the previous mathematical explanation. The most
superﬁcial answer is that, because the algorithm is about numbers and not
speciﬁcally about money, we should be able to oﬀer an explanation that is
valid in all contexts besides money. For example, why does this explanation
also explain the fact that 865 crabs + 32 crabs = 897 crabs, or 865 stars
+ 32 stars = 897 stars ? By the time you have found the answer, it is most
likely the case that you have also found a purely mathematical explanation
along the line of (16). On a deeper level, we have just seen how the math-
ematical explanation brings out issues that are hidden in the explanation
using money, such as place value and the basic laws of operations of ¸2 .
The mathematical explanation brings a deeper understanding not only of
the algorithm itself but also of these related issues, as we begin to see the in-
terconnectedness of these seemingly disparate concepts. Moreover, we want
an explanation that is suﬃciently robust to be applicable to all numbers big
or small. In this regard, it would be extremely clumsy to explain the addition
50, 060, 001 +870, 040 in terms of money, but relatively easy to do so by the
mathematical method above, as we shall see.
Let us now take up the issue of “carrying” from one column to the next.
From our point of view, this issue is no more than a minor reﬁnement of all
3 The Standard Algorithms 48
that has been said before. Consider 68 + 59:
6 8
5 9
+ 1 1
1 2 7
(17)
The diﬀerence from the previous situations is that, in the right column, the
sum of digits, i.e., 8 + 9, is no longer a single digit number but is 17. The
algorithm calls for carrying the tens digit “1” of 17 to the next column on
the left. This is indicated by the small 1 under 5, but in some other conven-
tions, the 1 is entered above 6 instead; such minor notational diﬀerences are
irrelevant to the mathematics under discussion. Then in adding the num-
bers in the tens column, this 1 is taken into account and result in the sum
6 + 5 + 1 = 12, which is again not a single digit number. In like manner
then, the “1” of 12 is carried to the (invisible) hundreds column. Because
there is no other number in the hundreds column, the last 1 is recorded in
the hundreds column and we get 127 as the ﬁnal sum.
The basic explanation is the same as before, so it suﬃces to consider only
the new features here. Let us begin with an explanation in terms of money.
Imagine that you have a stack of bills consisting of
6 ten-dollar bills, and
8 one-dollar bills,
and another stack consisting of
5 ten-dollar bills, and
9 one-dollar bills.
You decide to count the two stacks by counting the ten-dollar bills and the
one-dollar bills separately. Thus you have
11 (= 6 + 5) ten-dollar bills, and
17 (= 8 + 9) one-dollar bills.
But 17 one-dollar bills is the same as 1 ten-dollar bill plus 7 one-dollar bills.
So you trade 10 of your one-dollar bills for a ten-dollar bill and you now have
12 (= 11 + 1) ten-dollar bills, and
7 one-dollar bills.

(18)
3 The Standard Algorithms 49
But we can also trade in 10 of the 12 ten-dollar bills for 1 hundred dollar
bill, thereby getting
1 hundred-dollar bill,
2 ten-dollar bills, and
7 one-dollar bills.



(19)
This line of reasoning not only arrives at the same answer, which is 127, but
also exhibits the fact that (18) corresponds exactly to carrying the 1 to the
tens place in (17), while (19) corresponds to carrying the 1 to the hundreds
place in (17).
Finally, we give a purely mathematical explanation of (17). We preface
the explanation with the comment that, having given the detailed reasoning
in supported of the computation (13) citing the associative, commutative,
and distributive laws, we will not overwhelm you with such details from now
on but will instead ask you to be aware of their implicit presence at every
turn. With this understood,
68 + 59 = (6 10
1
+ 8 10
0
) + (5 10
1
+ 9 10
0
)
= (6 10
1
+ 5 10
1
) + (8 10
0
+ 9 10
0
)
Applying the distributive law, we get
68 + 59 = (6 + 5) 10
1
+ (8 + 9) 10
0
= (10 + 1) 10
1
+ (10 + 7) 10
0
= (1 10
2
) + (1 10
1
) + (1 10
1
) + (7 10
0
)
= (1 10
2
) + ([1 + 1] 10
1
) + (7 10
0
).
the last line explicitly shows how the 1 is carried to the next column to the
left, twice, in (16). So 68 + 59 = 127.
This is the right time and place to tie up a loose end. It was mentioned
after (12) that there is a reason for the algorithm to move from right to left.
The reason is one of economy of means : we want to simplify the algorithm
to the utmost. Consider the preceding problem of 68 + 59. Suppose we
want to embark on a diﬀerent addition algorithm by starting instead with
the leftmost column and move to the right. Then the ﬁrst step would be:
6 8
+ 5 9
1 1
3 The Standard Algorithms 50
Here we have carried the 1 to the hundreds column. In the next step, we add
the right column and get 17. This can be recorded in the next row again in
accordance with the place value of 17 thus:
6 8
+ 5 9
1 1
1 7
Then we add these two numbers and arrive at 127 as before:
6 8
+ 5 9
1 1
1 7
1 2 7
The ﬁrst comment is that this algorithm is certainly correct. Moreover,
the proper execution of this algorithm requires the same technique of car-
rying the 1 to the next column, the only diﬀerence being that, whereas the
carrying in (17) is recorded under 59, here it is done by using two extra
rows. Comparing the last computation with (17) reveals one virtue of (17):
it is compact, and this is the decisive factor. A beginner might welcome the
left-to-right algorithm at the beginning, but once familiarity sets in, (17) is
the algorithm of choice for most people. All the more so when the addition
involves numbers with many digits.
A passing comment should be made about the “unnaturalness” of making
children do things from right to left. One must put things in perspective. The
fact is that children learn to do many things that they consider unnatural.
Brushing teeth is one of them, and staying neat and clean is another. More-
over, inducing violent actions on a video screen by manipulating a joystick
is quite possibly one of the most unnatural things imaginable for children.
Yet, children all over the world welcome video or computer games with open
arms. Are we then to believe that they cannot learn to calculate from right
to left even after we explain to them why it is important that they do?
We have thus far discussed the addition algorithm in terms of concrete
numbers for the simple reason that it is very diﬃcult, notationwise, to discuss
this and other whole number algorithms using abstract notation. Neverthe-
less, we hope that the generality of the underlying mathematical reasoning
3 The Standard Algorithms 51
comes through so that, faced with other situations, you can both execute this
algorithm correctly and know how to justify it. Just to be sure, we will discuss
two more examples for further illustration. First, consider 165+27+83+829 .
(Recall that this makes sense without parentheses because of the associative
law.) Here is how the algorithm works out:
1 6 5
2 7
8 3
8 2 9
+ 1 2 2
1 1 0 4
(20)
The precise description of the procedure is nothing new: again we add
column-by-column, and move from right to left. Start with the right col-
umn and add: 5 + 7 + 3 + 9 = 24, so we carry the 2 to the next column
(to the left). Next, (6 + 2 + 8 + 2) + 2 = 20, so we again carry the 2 to the
next column. Finally, in the hundreds column: (1 + 8) + 2 = 11, and we
carry the 1 to the thousands column. Because there is no other number in
the thousands column, we record the 1 in the ﬁnal sum.
The explanation is the following:
165 + 27 + 83 + 829 = (1 10
2
) + (6 10
1
) + (5 10
0
)+
(2 10
1
) + (7 10
0
)+
(8 10
1
) + (3 10
0
)+
(8 10
2
) + (2 10
1
) + (9 10
0
)
= (9 10
2
) + (18 10
1
) + (24 10
0
)
But 24 10
0
= (2 10
1
) + (4 10
0
), so by the distributive law:
165 + 27 + 83 + 829 = (9 10
2
) + (18 10
1
) + (2 10
1
) + (4 10
0
)
= (9 10
2
) + ([18 + 2] 10
1
) + (4 10
0
)
This explains why we carried the 2 to the tens column in (20). Next, [18 +
2] 10
1
= 2 10
2
. Hence,
165 + 27 + 83 + 829 = (9 10
2
) + (2 10
2
) + (4 10
0
)
= ([9 + 2] 10
2
) + (4 10
0
)
This explains why we carried the 2 to the hundreds column in (20). In
any case, it is now clear that because [9 + 2] 10
2
= [10 + 1] 10
2
=
3 The Standard Algorithms 52
(1 10
3
) + (1 10
2
), we obtain:
165 + 27 + 83 + 829 = (1 10
3
) + (1 10
2
) + (4 10
0
) = 1104 ,
exactly as in (20).
As a ﬁnal example, let us use the addition algorithm to compute 50, 060, 001+
870, 040 and at the same time give the explanation.
5 0 0 6 0 0 0 1
8 7 0 0 4 0
+ 1
5 0 9 3 0 0 4 1
The explanation is simple:
50, 060, 001 + 870, 040 = ¦5 10
7
+ 6 10
4
+ 1 10
0
¦+
¦8 10
5
+ 7 10
4
+ 4 10
1
¦
= (5 10
7
) + (8 10
5
) + ([6 + 7] 10
4
)+
(4 10
1
) + (1 10
0
)
= (5 10
7
) + (8 10
5
)+
([10 + 3] 10
4
) + (4 10
1
) + (1 10
0
)
Look at the second and third terms of the right side: (8 10
5
) +([10 +3]
10
4
) = ([8 + 1] 10
5
) + (3 10
4
). This then accounts for carrying the 1 to
the column containing 8. Now, we obtain:
50, 060, 001 + 870, 040 =
(5 10
7
) + (9 10
5
) + (3 10
4
) + (4 10
1
) + (1 10
0
) .
As we mentioned above, an explanation of this addition problem using money
would be both clumsy and irrelevant.
Exercise 3.1 Explain to a fourth grader why the addition algorithm for
57032 + 2845 is correct, ﬁrst using the method of (16), then using money.
Exercise 3.2 Explain to a fourth grader why the addition algorithm
826 + 4907 is correct, with and without the use of money.
Exercise 3.3 Do the addition 67579+84937 both ways, from left to right,
and from right to left, and compare.
3 The Standard Algorithms 53
Exercise 3.4 Compute 123 +69 +528 +4 by the addition algorithm, and
give an explanation of why the computation is correct.
Exercise 3.5 Compute 7826+7826+7826 by the addition algorithm, and
give an explanation of why the computation is correct.
Exercise 3.6 Compute 670309+95000874 by the addition algorithm, and
give an explanation of why the computation is correct.
3.2 A subtraction algorithm
Subtraction has to be understood in terms of addition. Thus 37 − 19 is by
deﬁnition the number so that, when added to 19, yields 37. Thus:
(37 −19) + 19 = 37.
So adding 19 undoes the eﬀect of subtracting 19. Similarly, if m and n are
whole numbers, and m < n, then n−m is by deﬁnition the number so that,
when added to m , yields n. Thus,
(n −m) + m = n (21)
So again, adding m undoes the eﬀect of subtracting m. It also means that
in order to check whether a number x is equal to n − m, it is the same as
checking whether
x + m = n.
In particular, to verify a statement about subtraction, it suﬃces to verify
a statement about addition. This simple fact will be seen to be extremely
useful. On an intuitive level, it is common to think of 37 −19 as “taking 19
objects from 37 of them”, and n−m as “taking m objects from n of them”.
Activity: 1200 − 500 = ? 580, 000, 000 − 500, 000, 000 = ?
580, 000, 000 −20, 000, 000 = ? 15 10
6
−7 10
6
= ?
In terms of the number line, n −m has the following geometric interpre-
tation. Rewrite (21) as m + (n −m) = n (the commutativity of addition!).
Note that m and n are the lengths of the segments [0, m] and [0, n]. There-
fore, according to the interpretation of the sum of two whole numbers as the
length of a concatenated segment, the preceding equality implies that n−m
is exactly the length of the segment from m to n:
3 The Standard Algorithms 54
0 m n
' E
n −m
We can also look at (21) as is: (n − m) + m = n. This says that the
concatenation of [0, n − m] and [0, m] is a segment of the same length as
[0, n]. This means that we can arrive at the point n − m by going from n
to the left for a length of m:
0 n −m n
m
'
As is the case with the addition algorithm, the purpose of the standard
subtraction algorithm is to relieve the tedium of counting backwards 257
times from 658 in order to compute 658−257. The mathematics underlying
this algorithm (to be introduced presently) is so similar to that of the addition
algorithm that we can aﬀord to be brief. As before, we begin with the simple
case, e.g., 658 − 257, where the simplicity refers to the fact that each of
the digits in the ﬁrst number is at least as big as the corresponding digit in
the second number. The algorithm calls for lining up the digits of the two
numbers column-by-column from the right (as in the addition algorithm) and
then do subtraction of single digit numbers in each column, starting with the
right column and move left: 8 −7 = 1, 5 −5 = 0, 6 −2 = 4.
10
Schematically:
6 5 8
− 2 5 7
4 0 1
(22)
For the explanation of the algorithm, the following subtraction facts are
useful. Suppose l, m, n, a, b, and c are any whole numbers so that l − a,
m > b and n > c, we have:
(l + m + n) −(a + b + c) = (l −a) + (m−b) + (n −c) (23)
m(l −a) = ml −ma (24)
10
Again, the subtraction of 658−257 is reduced to three single-digit computations: 6−2,
5 −5, and 8 −7.
3 The Standard Algorithms 55
Equation (23) is entirely plausible if it is interpreted in terms of concrete
objects. For example, if you have three piles of oranges, having l, m and n
oranges in each pile, respectively, then taking a+b+c oranges away from the
three piles combined would leave behind the same total number of oranges
as taking successively a oranges from the pile of l oranges, b oranges from
the pile of m oranges, and c from the pile of n oranges. Equation (24) is a
variant of the distributive law (10) and can be made believable using oranges
in exactly the same way.
It is also instructive to directly prove (23) and (24), because we will get to
know the deeper meaning of the deﬁnition (21) in the process. First note
that (23) makes sense, i.e., it is in fact true that (l + m + n) > (a + b + c)
so that the left side of (23) makes sense. To see why, we make repeated use
of the third assertion of (11) near the end of ¸2 (to the eﬀect that A > B
and C > D imply A+C > B +D) and use the assumption of l > a, m > b,
and n > c to conclude that (l + m + n) > (a + b + c). Now to prove (23),
let x = l −a, y = m−b, and z = n −c, then the right side of (23) becomes
x + y + z, and we want to prove that
(l + m + n) −(a + b + c) = x + y + z.
According to the remark after deﬁnition (21), this is the same as checking
((a + b + c) + (x + y + z) = (l + m + n) (♣)
By the general associative law and general commutative law of addition, the
left side of (♣) is equal to (x + a) + (y + b) + (z + c), which can be further
simpliﬁed by use of
x + a = (l −a) + a = l
y + b = (m−b) + b = m
z + c = (n −c) + c = n
Thus the left side of (♣) is l + m + n, which shows that (♣) is true.
Simlarly, to show that (24) is true, let x = l − a. Then (24) becomes the
statement that mx = ml −ma. Again by the remark after (21), this would
be true if we can show
mx + ma = ml (♠)
But by the distributive law, the left side of (♠) is m(x +a) which in virtue
of x = l − a is equal to m([l − a] + a) = ml. This shows that (♠) is also
true. We have thus completed the formal proofs of (23) and (24)
3 The Standard Algorithms 56
It should be remarked that (23) is valid for more than three pairs of
numbers. For example, the analogue of (23) for ﬁve pairs of numbers would
read: if l > a, m > b, n > c, p > d, and q > e, then
(l + m + n + p + q) −(a + b + c + d + e) =
(l −a) + (m−b) + (n −c) + (p −d) + (q −e)
The proof of this more general version is of course the same as the case of
three pairs of numbers. A complete understanding of (23) and (24) must
await the introduction of negative numbers in Chapter 5.
Using (23) and (24), we can now give the explanation of (22):
658 −257 = (6 10
2
+ 5 10
1
+ 8 10
0
)−
(2 10
2
+ 5 10
1
+ 7 10
0
)
= (6 10
2
−2 10
2
) + (5 10
1
−5 10
1
)+
(8 10
0
−7 10
0
) (by (23))
= ([6 −2] 10
2
) + ([5 −5] 10
1
) + ([8 −7] 10
0
)
(by (24)
= 4 10
2
+ 1 10
0
= 401
We now tackle the general case which requires “trading”. Consider, for
example, 756 −389. The preceding column-by-column method breaks down
at the ﬁrst step because the subtraction 6 − 9 cannot be done as is using
whole numbers. In this case, the algorithm states:
Take 1 from the tens digit (which is 5) of 756 — thereby changing
the tens digit of 756 from 5 to 4 — and then do the subtraction in
the ones column, not as 6 −9, but as 16 −9 (which is 7). In the
tens column, we now have 4−8. Again, take 1 from the hundreds
digit (which is 7) of 756 — so that 7 becomes 6 — and do the
subtraction in the tens column as 14 − 8 (which is 6). Finally,
the subtraction in the hundreds column is now 6 −3 = 3.
Schematically:
6 4
,7 ,5 6
− 3 8 9
3 6 7
(25)
3 The Standard Algorithms 57
The explanation for (25) and the algorithm itself is the reverse of the explana-
tion given for carrying in the addition algorithm. We ﬁrst do it schematically
(note: the double arrow “ ⇐⇒ ” in the following means “is the same as”):
7 5 6
− 3 8 9
? ? ?
⇐⇒
700+ 50+ 6
− 300+ 80+ 9
?
⇐⇒
700+ 40+ 16
− 300+ 80+ 9
?
⇐⇒
600+ 140+ 16
− 300+ 80+ 9
?
⇐⇒
600+ 140+ 16
− 300+ 80+ 9
300+ 60+ 7
⇐⇒
7 5 6
− 3 8 9
3 6 7
Now, in greater detail, we use the associative law of addition repeatedly
to get:
756 = 700 + 50 + 6
= (600 + 100) + 50 + 6
= 600 + (100 + 50) + 6
= 600 + 150 + 6
= 600 + (140 + 10) + 6
= 600 + 140 + (10 + 6)
= 600 + 140 + 16
so that
756 −389 = (600 + 140 + 16) −(300 + 80 + 9)
= (600 −300) + (140 −80) + (16 −9) (by (23))
= 300 + 60 + 7
= 367
Note that the second line of the preceding calculation corresponds exactly
to the subtraction in (25). Note also that we avoided writing the number
756 and 389 in their expanded forms, but used a simpliﬁed version instead,
3 The Standard Algorithms 58
because the more complicated notation would have obscured the underlying
reasoning.
Although it is not a good policy in general to over-emphasize the role of
the laws of operations discussed in ¸2, it is nevertheless worthwhile to point
out the critical role played by the associative law of addition in the subtrac-
tion algorithm.
We add the usual remark that the mathematical reasoning behind the
preceding explanation is perfectly general and is applicable to any subtraction
a−b for any whole numbers a and b, so long as a is not smaller than b. Just
to be sure that this message gets across, we will work out another example:
5003 −465.
As before, the ﬁrst subtraction in the right column, 3 − 5, cannot be
carried out using whole numbers. We try to take a 1 from the tens digit
of 5003, which is unfortunately zero. So we go to the hundreds digit and
hope to take 1 from there. Again it is 0. This then requires going all the
way to the thousands digit “5” of 5003 to take 1 from 5. So 5 becomes
4 in the thousands digit of 5003, but the hundreds digit of 5003 becomes
10 +0 = 10 . From this 10, we can take 1 to bring it down to the tens digit.
in the process, the hundreds digit becomes 9 (instead of 10), and the tens
digit is now 10+0 = 10 . We are now back to our original problem of getting
1 from the tens digit to change the 3 to 10 + 3 = 13 . We do so, and the
tens digit becomes 9. So now, the subtraction in the ones column becomes
13 − 5 = 8 . The subtraction in the tens column becomes 9 − 6 = 3 , and
that in the hundreds digit becomes 9 − 4 = 5 . The thousands digit is of
course 4. Schematically:
4 9 9
,5 ,0 ,0 3
− 4 6 5
4 5 3 8
3 The Standard Algorithms 59
The explanation using the associative law of addition follows (this time,
we will dispense with the use of parentheses altogether):
5003 = 5000 + 3
= 4000 + 1000 + 3
= 4000 + 900 + 100 + 3
= 4000 + 900 + 90 + 10 + 3
= 4000 + 900 + 90 + 13
So using the generalized version of (23) for four pairs of numbers, we have:
5003 −465 = (4000 + 900 + 90 + 13) −(0 + 400 + 60 + 5)
= (4000 −0) + (900 −400) + (90 −60) + (13 −5)
= 4000 + 500 + 30 + 8
= 4538
We note that the subtraction algorithm is again one that works from right
to left. Just as in the case of the addition algorithm, one can work from left
to right, but the amount of backtracking needed for making corrections is
even greater here than in the case of addition.
Activity: Do the preceding subtraction 5003 −465 from left to
right, and compare with the computation from right to left.
It is worth repeating that there is absolutely nothing unnatural about teach-
ing children to do something from right to left.
For special numbers, there are usually tricks to make computations with
them much more pleasant. This applies in particular to the subtraction
algorithm. Let us give one such example: the preceding problem 5003−465
can be done very simply as follows:
5003 −465 = 4 + 4999 −465 = 4 + 4534 = 4538 .
Similarly,
30024 −8697 = 25 + 29999 −8697 = 25 + 21302 = 21327.
3 The Standard Algorithms 60
The point here is that the subtractions with rows of 9’s in the ﬁrst number
can be done without trading and can therefore be done easily by mental
math. One can even do it from left to right if one wishes. The same trick
can be used for any subtraction problem in which the ﬁrst number is close to
a multiple of 10
n
, where n is any whole number. Then by changing the ﬁrst
number to a small number plus another one with a row of 9’s, the subtraction
can be done with no eﬀort.
The preceding algorithm for subtraction when the ﬁrst number is close
to a multiple of 10
n
is so striking that it would have spoiled the fun if
we had called attention to a technical point. Nevertheless, mathematics
must be served, if only discretely this time, so we attend to this technical
point here. Take, for instance, 5003 − 465. The point is that the step
4 + 4999 − 465 = 4 + 4534 actually involves the associative law for integers
(positive and negative whole numbers, see Chapter 5). Precisely, we have
(4 + 4999) −465 = (4 + 4999) + (−465)
= 4 + (4999 + (−465))
= 4 + 4534
The reason why the preceding is worthy of a separate discussion is of course
the fact that the associative law in its naive form is not always valid for
subtraction, e.g., (19 − 4) − 5 ,= 19 − (4 − 5). The mystery surrounding
the associative law for subtraction would be dispelled once we do integer
operations correctly. See Chapter 5.
As a ﬁnal note on the subtraction algorithm, there is a natural alterna-
tive algorithm if negative numbers — together with a few simple arithmetic
properties — are allowed to be used. To illustrate, consider the previous
problem of 756 − 389 : We continue to do column-by-column single digit
subtractions, but now we can use negative numbers to record the diﬀerences
and we are free to do the column-by-column subtractions in any order, from
left to right or right to left. So:
7 5 6
− 3 8 9
4 [[−3]] [[−3]]
where each [[−3]] indicates the result of the column subtraction in that col-
umn. To get the ﬁnal answer, the algorithm says:
3 The Standard Algorithms 61
treat 4[[−3]][[−3]] as if it were a whole number with digits 4, [[−3]],
and [[−3]], and write it out in expanded form.
Thus:
4[[−3]][[−3]] = (4 10
2
) + ((−3) 10
1
) + ((−3) 10
0
)
= 400 −30 −3
= 370 −3 = 367 .
Notice that the subtractions here are much more tractable than those in the
original.
The explanation is simple enough and is based on (23) and (24):
756 −389 = ¦(7 10
2
) + (5 10
1
) + (6 10
0
)¦−
¦(3 10
2
) + (8 10
1
) + (9 10
0
)¦
= ([7 −3] 10
2
) + ([5 −8] 10
1
) + ([6 −9] 10
0
)
= 400 + ((−3) 10) + (−3) = 367
There are pros and cons regarding which of two algorithms is “better”.
The ﬁrst is simpler, but the second may be less prone to computational er-
rors, at least if the user is ﬂuent with negative numbers. Because teaching
third graders about the most elementary aspects of negative numbers is a
realistic goal, the second algorithm should be a viable option in schools.
Exercise 3.7 Give an interpretation of (22) in terms of money.
Exercise 3.8 Explain to a fourth grader why the subtraction algorithm
for 563 −241 is correct, with or without money.
Exercise 3.9 Give an interpretation of (25) in terms of money.
Exercise 3.10 Explain to a fourth grader why the subtraction algorithm
for 627 −488 is correct, with and without the use of money.
Exercise 3.11 (a) Use the subtraction algorithm to compute 2403 −876
and explain why it is correct. (b) Do the same with 76431 −58914.
Exercise 3.12 Compute 800, 400 − 770, 982 in two diﬀerent ways, and
explain why what you have done is correct.
Exercise 3.13 Compute 26, 004−8325 two ways, once using the standard
algorithm, and once using the preceding “negative number” algorithm.
Exercise 3.14 Let a, b, c be whole numbers. (a) Prove that a +b < c is
the same as a < c − b. (b) Suppose c < a and c < b. Prove that a < b
is the same as a −c < b −c.
3 The Standard Algorithms 62
Exercise 3.15 Find shortcuts to compute the following: 8 875 = ?
9996 25 = ? 103 97 = ? 86 94 = ?
3.3 A multiplication algorithm
We next take up the question of how to compute the product of two numbers
such as 826 73 without having to add 73 to itself 826 times. Bearing in
mind the leitmotif enunciated at the beginning of the section, we proceed to
break up the computation into a series of computations involving one digit
at a time. In this case, the distributive law does the breaking up:
826 73 = 826 (7 10 + 3) (26)
= (826 7) 10 + (826 3)
Thus
the multiplication (of 826) by a multi-digit number 73 has been
reduced to two simpler computations: multiplication by 7 (i.e.,
826 7) and multiplication by 3 (826 3), as given by (26).
We now further break up these two tasks into yet simpler tasks.
Let us ﬁrst look at 826 3. Instead of adding 3 to itself 826 times, we
apply the distributive law one more time:
826 3 = ¦(8 10
2
) + (2 10) + 6¦ 3 (27)
= (8 3) 10
2
+ (2 3) 10 + (6 3)
where of course we have also made use of the associative and commutative
laws of multiplication to conclude that, e.g., (8 10
2
) 3 = 8 3 10
2
. See
the discussion in ¸2 above equation (10). Thus 826 3 will be computable
according to (27), as soon as we know the products of single-digit numbers:
83, 23, and 63. This is why a ﬂuent knowledge of the multipli-
cation table is essential, because it lies at the heart of all multiplication
problems. In any case, from (27) we obtain:
826 3 = (24 10
2
) + (6 10) + 18.
3 The Standard Algorithms 63
Experience with the addition and subtraction algorithms tells us that we
should proceed by working from right to left: 18 = 10 + 8, so that
826 3 = (24 10
2
) + ([6 + 1] 10) + 8 (28)
= (24 10
2
) + (7 10) + 8
Now 24 10
2
= ([2 10] + 4) 10
2
= (2 10
3
) + (4 10
2
), so that
826 3 = (2 10
3
) + (4 10
2
) + (7 10) + 8 (29)
Equations (27)–(29) explain the following multiplication algorithm when one
number is single-digit:
8 2 6
3
2 1
2 4 7 8
(30)
The precise description of the algorithm for the multiplication of 826 by a
single digit number 3 is this. Multiply each digit of 826 by 3, from right to
left: 3 6 = 18, so carry the 1 to the tens column; 3 2 = 6, so the tens
digit of the answer is 6 + 1 = 7; 3 8 = 24, so carry the 2 to the hundreds
column and get 24 for the thousands and hundreds columns respectively.
To make sure that the multiplication algorithm with a single-digit multi-
plier is clearly understood, we will quickly do another example, 826 7, and
give a brief explanation.
8 2 6
7
5 1 4
5 7 8 2
It yields the right answer because:
826 7 = ¦(8 10
2
) + (2 10) + 6¦ 7
= (56 10
2
) + (14 10) + 42
= (56 10
2
) + (14 10) + ([4 10] + 2)
= (56 10
2
) + ([14 + 4] 10) + 2 .
The last line explains the carrying of the 4 to the tens column. Then:
826 7 = (56 10
2
) + (18 10) + 2
= (56 10
2
) + (10
2
+ [8 10]) + 2
= ([56 + 1] 10
2
) + (8 10) + 2 ,
3 The Standard Algorithms 64
and the last line explains the carrying of the 1 to the hundreds column.
Recall that we are looking at the multiplication algorithm with a single-
digit multiplier because (26) reduces the general case to this special case. It
is time to return to the general case, say (26). We now know:
826 7 = 5782
826 3 = 2478 .
According to (26) then,
826 73 = (5782 10) + 2478
= 2478 + 57820 ,
so that schematically we have:
8 2 6
7 3
2 4 7 8
+ 5 7 8 2 0
6 0 2 9 8
Because we are used to treating an empty slot as a zero (cf. the discussion
of the addition algorithm right above (12), for example), it is customary to
omit the “0” at the end of 57820 and just write:
8 2 6
7 3
2 4 7 8
+ 5 7 8 2
6 0 2 9 8
(31)
We are in a position to summarize this algorithm as follows:
To compute say 826 73, take the digits of the second factor 73
individually, compute the two products with single digit multiplier
— i.e., 826 3 and 826 7 — and, when adding them, shift the
one involving the tens digit (i.e., 7) one digit to the left, as in
(31).
3 The Standard Algorithms 65
This is commonly called the standard multiplication algorithm. Other varia-
tions are possible. For example, one alternative algorithm is
8 2 6
7 3
5 7 8 2
+ 2 4 7 8
6 0 2 9 8
Essentially, this means we run the algorithm from left to right, ﬁrst multi-
ply by 7 before mutiplying by 3. Mathematically, we do not consider such
formal diﬀerences to be a diﬀerence at all. It is worth noting that even the
algorithm with a single-digit multiplier can be carried out from left to right.
For example, 6718 5 can be done this way:
6 7 1 8
5
3 0
3 5
5
+ 4 0
3 3 5 9 0
Activity: Give a precise explanation of the preceding algo-
rithm.
Pedagogical Comment: In a classroom, the most salient feature of
this algorithm that catches students’ attention may well be the shifting of
the product involving the tens digit (i.e., 826 7) to the left by one digit.
This should be carefully explained to them in terms of place value: we are
actually looking at 82670 and the shifting of digit is caused by the presence
of the “0” in the ones digit. End of Pedagogical Comment.
In order to ensure that the generality of the preceding reasoning behind
the algorithm is understood, let us brieﬂy explain how to do a more compli-
3 The Standard Algorithms 66
cated problem.
5 2 7
3 6 4
2 1 0 8
3 1 6 2
+ 1 5 8 1
1 9 1 8 2 8
(32)
This is because:
527 4 = 2108
527 6 = 3162
527 3 = 1582 ,
and so by the distributive law:
527 364 = 527 (3 10
2
+ 6 10 + 4)
= (527 3) 10
2
+ (527 6) 10 + (527 4)
= 158100 + 31620 + 2108
The last line explains the vertical alignments of the digits of the three prod-
ucts 2108, 3162, and 1581 in (32). In particular, because the 3 of 364 is the
hundreds digit, the product 5273 is shifted two digits to the left in keeping
with the fact that it is really 527 300 that we are adding.
Exercise 3.16 Explain to a 4th grader why the multiplication algorithm
for 86 37 is correct.
Exercise 3.17 (a) Which 2-digit number, when multiplied by 89, gives
a 4-digit number that begins and ends with a 6? (b) List all the 3-digit
numbers which have the following properties: the sum of the digits is 12,
and when multiplied by 15 they give a 5-digit number which ends with a 5
(i.e., the ones digit is 5). (Clearly this problem can be done by guess and
check. You are however asked to use reasoning to quickly dospatch it by
narrowing down the choices.)
Exercise 3.18 Use the multiplication algorithm to compute
1 8
5 0 0 0 9 2
?
3 The Standard Algorithms 67
and explain why it is correct. Compare with the same algorithm applied to
5 0 0 0 9 2
1 8
?
Exercise 3.19 Compute 4208 879 by the multiplication algorithm and
explain why it is correct.
3.4 Division-with-remainder
When asked to divide 23 by 4, we all know the answer: the quotient is
5 and the remainder is 3. In general, given whole numbers a and b with
b ,= 0 , we likewise want to know what the quotient and remainder are when
a is divided by b. This knowledge is critical not only for the long division
algorithm of ¸3.5 below but also for the discussions of fractions in Chapter 2
and decimals in Chapter 4. Before we can come up with an answer, however,
we need to know precisely what the “quotient” of a ÷ b means and what
the “remainder” means. Most school textbooks do not deem it necessary to
explain these concepts, or if they do, they typically say the following:
division An operation on two numbers that tells how many
groups or how many in each group.
quotient The answer in division.
remainder The number that is left over after dividing.
If you look more closely at the text proper, you would ﬁnd the statement
that the remainder should be less than the “divisor” b, but this still leaves
out a clear statement about what a “quotient” is and what “left over after
dividing” means. Such vagueness would not serve any purpose because unless
the student already knows what “division” means, the above explanations
give no information. For example, how do we compute the “quotient” of
6810255956001 ÷ 28747 if we do not know its precise meaning? The usual
attempt at an explanation of division would mention taking away multiples
of 28747 until “the remainder” is “smaller than 28747”. This unfortunately
begs the question of what a “quotient” is and whether a negative number
may be considered to be “smaller than” 28747.
3 The Standard Algorithms 68
As teachers, we want to convey the clear message to our children that in
mathematics, no guesswork is needed for its mastery. We want to let them
know that it is an open book that everybody can read. Among all branches of
knowledge, mathematics is characterized by its WYSIWYG quality — what
you see is what you get — and you have no need to assume anything that
is not already explicitly stated. This is another way of expressing the fact
that every conclusion we draw in mathematics depends completely on what
is stated explictly up front. This is what we have been doing so far, and we
intend to continue doing it for the rest of this monograph.
To return to division, let us ﬁrst ﬁx the meaning of a ÷b where a and
b are whole numbers and b ,= 0 . (We will always assume b ,= 0 in this
situation because we do not want to divide by zero; see the discussion in ¸4
below.) Generally speaking, division is to multiplication as subtraction is to
addition: one undoes the other. However, there are certain wrinkles to this
simple-minded statement, and we will be careful to address them.
Recall that multiplication is repeated addition: by deﬁnition,
qb = b + b + + b

q
(see (2) of ¸1). Division a ÷ b is roughly speaking repeated subtraction, but
the precise meaning of this phrase requires a rather long-winded explanation.
First of all, if m is a whole number, the product mb is called a multiple of
b, or more precisely, the m-th multiple of b. (In particular, 0 is a multiple of
b, by deﬁnition because 0 = 0 b.) Intuitively, what we are going to do is
to repeatedly subtract b from a until we get to the point where the next
subtraction will not be possible because what is left is smaller than b (recall:
for x − b to make sense, we must have x ≥ b). Symbolically though, this
way of doing things is very awkward, so we do something which is the same
11
but which is easier to express. In greater detail, what we do is to take (i.e.,
subtract) successive multiples of b from a: a −0, a −b, a −2b, a −3b, a −4b,
. . . , until eventually we come to a multiple qb of b so that the next multiple
— which is (q +1)b — exceeds a. In symbols, q is that whole number so that
a ≥ qb but a < (q + 1)b. (33)
11
An explanation of why it is the same is given in the next ﬁne-print indented passage.
3 The Standard Algorithms 69
By deﬁnition, we call this q the quotient of a÷b. It follows that the quotient
q is the largest multiple of b that one can take away from a, because we cannot
perform the next subtraction a−(q +1)b for the reason that a < (q +1)b, by
(33). It is intuitively clear, and we shall prove precisely below, that after we
have taken q multiples of b’s from a, what is left behind is less than b. We
call a −qb the remainder of a ÷b. To complete the terminology, we call b
the divisor and a the dividend of the division a ÷b.
We now proceed to bring out a critical property of the remainder that
was alluded to above, namely, a − qb < b. We ﬁrst give a numerical proof,
but presently we will also give a pictorial proof that makes the reasoning
perfectly obvious. Here then is the numerical proof: from (33) we have
a < (q + 1)b. By taking qb away from both sides, the inequality does not
change (see Exercise 3.14(b)). So we get
a −qb < (q + 1)b −qb
= ¦(q + 1) −q¦b (by (24) of ¸3.2)
= b
So we get a − qb < b. Incidentally, we also know from (33) that a ≥ qb,
so a − qb ≥ 0. We may therefore summarize the preceding two facts in the
following double inequality:
the remainder a −qb satisﬁes 0 ≤ a −qb < b (34)
Note that the left inequality sign of (34) is a weak inequality (i.e., allowing for
equality) because the remainder does equal 0 sometimes, e.g., when a = qb.
For example, if a = 23 and b = 4, then the multiples of 4 are 4, 24 = 8,
3 4 = 12, 4 4 = 16, 5 4 = 20, 6 4 = 24, and we stop at the 6th
multiple of 4 because already
23 ≥ (5 4) but 23 < (6 4).
Therefore the quotient of 23 ÷4 is 5, and the remainder is 23 −(5 4) = 3.
If we take a = 12 and b = 3, however, then we get the happy coincident that
12 = 4 3, i.e., 12 is exactly the 4th multiple of 3. In this case, the quotient
of 12 ÷3 is 4, with remainder 0.
If a = qb for some whole number q, we say b divides a. In symbols: b[a.
Note that “ b divides a” says exactly the same thing as “ a is a multiple of
b”.
3 The Standard Algorithms 70
Activity: Find the quotient and remainder in each of the fol-
lowing divisions by listing the multiples of the divisor (but do
not use “long division”, whatever that means): 33 ÷7, 46 ÷9,
98 ÷19, 188 ÷37.
We now give the pictorial representation of the quotient and remainder
using the number line. Consider 27 ÷6. The multiples of 6 are 0, 6, 12, 18,
24, 30, etc., as shown.
0 6 12 18 24 30
27
c
The picture clearly displays the fact that 27 is trapped between the two
multiples of 6: 24 and 30. The remainder of 27 ÷6 is just the length of the
segment between 27 and 4 6 = 24, which is clearly less than the length
between 24 and 30. Therefore the remainder 27 − (4 6) < 6 (= 30 − 24).
Thenquotient is 4 because 24 = 4 6.
As another example, consider 97 ÷12. The multiples of 12 are: 0, 12, 24,
36, 48, 60, 72, 84, 96, 108.
0 12 24 36 48 60 72 84 96 108
c
97
Again, the dividend 97 is trapped between the two multiples 96 and 108
of 12, and the remainder of 97 ÷ 12 is the length of the segment [96, 97]
(the segment between 96 and 97), which is less than 6 (= the length of the
segment [96, 108]). Because 96 = 8 12, the quotient of 97 ÷12 is 8.
In general, we have a ÷ b, then the multiples of b are equally spaced
markers (= points) on the number line, b units apart. The whole number
a has to be trapped between two of these multiples, or right at one of the
multiples. In the former case, let a be between qb and (q + 1)b.
c
a
0 b 2b 3b qb (q + 1)b
3 The Standard Algorithms 71
Then, the remainder is just the length of the segment [qb, a], and it is clearly
smaller than the length of the segment [qb, (q + 1)b] (= b).
On the other hand, if a is at one of the multiples of b, let it be at qb, as
shown.
c
a
0 b 2b 3b (q −1)b qb
In this case, a = qb and of course the remainder is 0.
We can summarize our discussion in the following theorem.
Division-with-Remainder. Given any two whole numbers a and b,
with b > 0, there exist a whole number q, called the quotient of a ÷b, so that
the remainder a −qb satisﬁes (34), i.e.,
0 ≤ a −qb < b.
This theorem is more commonly cast in a diﬀerent form, as follows. Let
the remainder a − qb be denoted by r, then a − qb = r by deﬁnition, which
can be rewritten as a = qb + r. The condition (34) now says 0 ≤ r < b.
Hence, we have an equivalent formulation of Division-with-Remainder:
Division-with-Remainder (Second Form). Given any two whole
numbers a and b, with b > 0, there exist a whole number q, called the quotient
of a ÷b, and a whole number r, called the remainder of a ÷b, so that
a = qb + r where r satisﬁes 0 ≤ r < b. (35)
An added remark about the division-with-remainder will be relevant in
the discussion of long division (¸3.5) and the decimal expansion of a fraction
(¸4 of Chapter 4). Ordinarily when one divides a by b, there is an implicit
assumption that a is bigger than b. However, in the above statement of
division-with-remainder, the relative sizes of a and b are irrelevant. For ex-
ample, for 5 ÷ 32, we have 5 = (0 32) + 5, so that the quotient is 0 and
the remainder is 5. For 29 ÷127, we have 29 = (0 127) +29, with quotient
0 and remainder 29. The point is that the division-with-remainder makes
sense for any a ÷b, so long as b > 0.
3 The Standard Algorithms 72
We now tie up two loose ends left dangling in the preceding discussion.
The ﬁrst is the explanation of why “repeated subtraction of b from a” is the
same as ”subtracting successive multiples of b from a”. The second is the the
omission of the uniqueness of the quotient in the division-with-remainder.
Repeated subtractions of b from a means of course (a −b), ((a −b) −b),
(((a − b) − b) − b), . . . . We used instead a − b, a − 2b, a − 3b, . . . , in
the above. What needs to be pointed out is that ((a − b) − b) = a − 2b,
(((a −b) −b) −b) = a −3b, and in general
( ((a −b) −b) −b

q
) = a −qb (†)
for all whole numbers q. Of course (†) is intuitively obvious, because whether
one takes b away from a one at a time, q times in succession, or take qb
away from a all at once, what is left behind should be the same, However,
it is also important to realize that intuition need not be the sole arbiter
of mathematical truths, so we shall give a precise proof of (†) which also
happens to be instructive.
We begin with an observation: if A, B are whole numbers so that (a −
M) −N ≥ 0, then
(a −M) −N = a −(M + N) (‡)
Here is the reason: Let x = (a −M) −N. Then (‡) becomes the statement
that x = a − (M + N). Therefore, to prove (‡), we only need to prove
x = a − (M + N), and for this, — we recall the remark made after (21) in
¸3.2 — it suﬃces to prove x + (M + N) = a. To this end, observe again as
a consequence of the deﬁnition of subtraction in (21) that x = (a −M) −N
means
x + N = a −M. ()
So we have:
x + (M + N) = x + (N + M) (commutative law of +)
= (x + N) + M (associative law of +)
= (a −M) + M (by ())
= a (by deﬁnition of subtraction in (21))
This proves x + (M + N) = a, and therewith (‡).
We can now prove (†) in succession for q = 1, q = 2, q = 3, . . . , etc. If
q = 1, (†) merely says a −b = a −b. Let q = 2, then (†) states that
[a −b] −b = a −2b, ()
but this follows directly from (‡) by letting M = N = b. Next let q = 3.
Then we must prove:
([a −b] −b) −b = a −3b. ()
3 The Standard Algorithms 73
This is true because:
([a −b] −b) −b = (a −2b) −b (by ())
= a −(2b + b) (by (‡) with M = 2b and n = b)
= a −3b.
Now let q = 4. Then we have to prove
(([a −b] −b) −b) −b = a −4b,
and this is true because
(([a −b] −b) −b) −b = (a −3b) −b (by ())
= a −(3b + b) (by (‡) with M = 3b and n = b)
= a −4b,
as desired. The next thing to prove is the case of (†) for q = 5, etc. But it
is clear by now, with the pattern of proof ﬁrmly established, that rest of the
argument would proceed would proceed in similar fashion. Therefore (†) is
true for all whole numbers q.
We next turn to the division-with-remainder and point out that the
quotient q of the theorem is actually unique. What this means is that suppose
we have 31 ÷7. Then we know the quotient is 4, and 0 ≤ 31 −(4 7) < 7.
Now suppose there is another whole number s so that 0 ≤ 31 −(s 7) < 7.
The theorem implies that this s must be equal to 4 too. The reason for this
is based on a general fact: if we have two whole numbers m and n so that
0 ≤ m, n < 7, — and we may as well assume m ≤ n — then it is quite clear
that the diﬀerence n−m is a whole number which is not a nonzero multiple
of 7. See the picture:
0 m n 7
We now apply this simple observation to 31÷7 by letting m = 31−(47) and
n = 31−(s7). As before, we may assume m ≤ n. Then we know that m−n
is not a nonzero multiple of 7. Using the simple fact that (l −a) −(m−b) =
(l −m) + (b −a), to be proved in Chapter 5, we have:
m−n = ¦31 −(4 7)¦ −¦31 −(s 7)¦
= (31 −31) +¦(s 7) −(4 7)¦
= (s −4) 7
Because we know ahead of time that m−n is a whole number, we see that
(s −4) 7 must be a whole number, and as such, it has to be a multiple of
3 The Standard Algorithms 74
7. We already know that it cannot be a nonzero multiple of 7, so it can only
be the zero multiple, in which case s −4 = 0, which is to say, s = 4. Exactly
as claimed.
In general, the quotient q of a÷b is unique in the sense that if s is another
whole number so that also 0 ≤ a −sb < b, then necessarily s = q. The proof
is identical to the preceding argument as soon as the numbers 31, 7 and 4
are repalced by a, b, and q, respectively. We will not repeat the argument.
Moreover, if the quotient is unique, then so is the remainder a −qb.
Conceptually, the uniqueness is important because we talk freely about
the quotient of a division and the remainder of a division. So without being
aware of it, we tacitly assume the uniquness in question. More is true.
The requirement that both q and r be whole numbers is critical
to their uniqueness.
For example, the ﬁrst example 23 ÷ 4 of this subsection has (as we know)
quotient and remainder equal to 5 and 3, respectively. However, if we are
allowed to use fractions, for example, then we could have
23 = 5 +
1
4
4 + 2 and 0 ≤ 2 < 4
(Although we have not yet taken up the subject of fractions, there is no harm
in using them for illustration.) This would give a “quotient” of 5 +
1
4
and a
“remainder” 2. In fact, we can write many such equations at will, e.g.,
23 =
22
4
4 + 1 and 0 ≤ 1 < 4
This should serve as reminder how delicate the uniqueness of the quotient and
remainder really is. In school mathematics, however, such subtlety is usually
glossed over. While one cannot say that such negligence does elementary
school mathematics great harm, we hope nevertheless to have convinced you
that, as a teacher, you should be aware of it.
The special case of division-with-remainder where the remainder is 0 oc-
cupies a place of distinction, and we proceed to discuss it in some detail. So
let a = qb, where q is the quotient. In this case, it is customary to write
q = a ÷b. It follows that if a = qb, then
(a ÷b) b = a and (ab) ÷b = a (36)
(It is of some value to point out that the second assertion is strictly a conse-
quence of the deﬁnition: let x = ab, then by the deﬁnition above, a = x ÷b,
which is another way of writing a = (ab) ÷b.) The equations in (36) clearly
display division and multiplication as two operations that undoes each other.
3 The Standard Algorithms 75
If q = a ÷ b, the fact that a = qb = b + b + + b (q times) means that
if we take b objects from a each time and do it q times, we would exhaust a.
Recalling once again that q = a ÷b, we have
a ÷b is the total number of groups when a objects are partitioned
into equal groups of b objects
This is called the measurement interpretation of division (in case there is no
remainder). However, there is another common way in which we use division.
Activity: In a third grade textbook, division is introduced as
follows:
You can use counters to show two ways to think about
dividing.
(A) Suppose you have 18 counters and you want to
make 6 equal groups. You can divide to ﬁnd how
many to put into each group.
(B) Suppose you have 18 counters and you want to
put them into equal groups, with 6 counters in each
group.
Although you can see easily in this special case that the answer
to both problems is 6, discuss which of these two divisions uses
the measurement interpretation and which requires a new under-
standing of division. It may help you to think more clearly if we
replace 18 by 4023, and 6 by 27 in the above problem.
Suppose as usual we have a objects and q = a÷b. Suppose we divide the
a objects into b equal groups, how many are in each group? Let a = 15 and
b = 5 so that 3 = 15 ÷5. We know that 15 can be partitioned into 3 groups
of 5’s, and we represent it pictorially by dots as follows:
• • • • •
• • • • •
• • • • •
Now if we count the dots by columns, we get 5 groups of 3’s and 15 =
3 + 3 + 3 + 3 + 3 = 5 3. Thus if we divide 15 (= a) objects into 5 (= b)
equal groups, there will be 3 objects in each group. But of course, 3 is just
3 The Standard Algorithms 76
the quotient of 15÷5, which by deﬁnition is equal to 35 (= qb. Comparing
with 15 = 3 + 3 + 3 + 3 + 3 = 5 3, we immediately recognized that the
commutativity of multiplication 3 5 = 5 3 is at work.
In general then, with a objects and q = a ÷ b, then a = qb. By the
commutativity of multiplication, a = qb = bq = q + q + + q (b times), so
that if we divide a objects into b equal groups, there will be q objects in each
group. This leads to the following: assuming q = a ÷b, then
a ÷ b is also the number of objects in each group when a objects
are divided into b equal groups.
This is called the partitive interpretation of division (in case there is no re-
mainder). The fact that this interpretation is valid (i.e., yields the same
number as the measurement interpretation) is due to the fact that multipli-
cation among whole numbers is commutative. For example, in the preceding
Activity, task (A) requires that you use the partitive interpretation of division
while task (b) requires the measurement interpretation.
At the risk of pointing out the obvious, both meanings of division are
common place in everyday life. Suppose you give a party and you make a
bowl of punch. If you want to ﬁnd out how many cups of punch there are in
the bowl, you are making a measurement division of the amount of ﬂuid in
your punch bowl by the amount of ﬂuid in your cup. On the other hand, if
four people decide to drink up the whole bowl of punch but wish to exercise
caution by ﬁrst computing how much ﬂuid each must be prepared to take
in if each drinks an equal amount, then these people will be doing partitive
division of the amount of ﬂuid in your punch bowl by 4. Everywhere you
look, you will ﬁnd both kinds of division done around you.
Example There is no better illustration of the two meanings of division
than the problem of traveling (or motion). Suppose a car goes from town
A to town B at a constant speed, which means that the distance traveled
within any one-hour time interval is always a ﬁxed constant. (Warning to
the reader: while this may seem like a good example of contextual learning,
we should not delude ourselves into believing that this is anywhere close to a
“real world” situation. Drastic oversimpliﬁcations are involved. For instance,
one rarely manages to drive at a constant speed for more than a few minutes
in real life. There is also the implicit idealization in that the car is driving
on a freeway that connects the two towns in a straight-line. How often does
this happen in everyday driving?) A typical question is then the following:
3 The Standard Algorithms 77
suppose the distance between towns A and B is 264 miles, and the car gets
to town B after 4 hours, what is the speed? If 264 miles is covered in 4
hours, we just partition 264 into 4 equal parts and the number of miles in
one part is the number of miles traveled n one hour (“equal parts” because
speed is assumed constant). This is the partitive meaning of the division
264
4
, which is 66. So the speed is 66 miles per hour. Another typical question
is the following: suppose the speed is a constant 58 miles per hour and the
distance between A and B is 522 miles, how many hours does it take to go
from A to B? Here, we know that the car would be 58 miles from A after
1 hour, 58 + 58 = 116 miles from A after 2 hours, 58 + 58 + 58 = 174 miles
from A after 3 hours, etc. So the question becomes how many 58’s there are
in 522. This is the measurement interpretation of
522
58
, which is 9. So it
takes 9 hours to go from A to B.
To summarize:
For motion in constant speed, computing the speed when distance
and time are given is a partitive division problem, while comput-
ing the time to travel a certain distance at a given speed is a
measurement division problem.
Finally, we give another geometric interpretation of division without re-
mainder. In ¸2, we introduced an area model for multiplication. According
to this model, 2 3, for example, would be modeled as the area of the
rectangle with vertical side equal to 2 and horizontal side equal to 3:
Now suppose we ask for 6 ÷ 3 = ? From the point of view of the area
model, this means we have a rectangle with area equal to 6 and a horizontal
side equal to 3, and we want to know what the length of the vertical side is:
area = 6
3
' E
?
3 The Standard Algorithms 78
Similarly, the division 35 ÷7 may be interpreted as asking for the length
of the vertical side of a rectangle with area 35 and with the horizontal side
equal to 7:
area = 35
7
' E
?
Of course, for whole numbers, such a geometric interpretation of division
is no more than slightly entertaining. However, when we come to the division
of fractions, the geometric interpretation would acquire added signiﬁcance.
Exercise 3.20 Is 24 the quotient of 687÷27? Is 13 the quotient of 944÷46
? Explain. (No calculator allowed.)
Exercise 3.21 Is 6977 the remainder of 124968752 ÷ 6843 ? Why? (No
calculator is allowed.)
Exercise 3.22 . By taking multiples of the divisor, ﬁnd the quotient and
remainder in each of the following cases: 964 ÷ 31, 517 ÷ 19, 6854 ÷ 731,
4972086 ÷873, and 4972086 ÷659437.
Pedagogical Comment: The use of (only) a four-function calculator
for the last two items involving 4972086 is allowed. However, it would be
instructive to ﬁrst ask for a ballpark ﬁgure of the quotient without the use of
a calculator. This is a good exercise in making estimates, and would save a
lot of guess-and-check in getting the correct quotient in each case. It should
also be mentioned that there is an eﬀective way to use the (four-function)
calculator to get the quotient and remainder without any trial and error.
How to do this and why it is true should lead to an interesting classroom
discussion (one that in fact presupposes some knowledge of decimals). End
of Pedagogical Comment.
3 The Standard Algorithms 79
Exercise 3.23 Let r be the remainder of a ÷ b. Suppose a = mA and
b = mB for some whole numbers m, A and B. Let R be the remainder
of A ÷ B. What is the relationship between R and r? Give a detailed
explanation of your answer. (Caution: This problem is deceptive because
it seems almost trivial, but the explanation is actually quite subtle and it
requires the use of the uniquness of both the quotient and remainder which
is discussed in the indented ﬁne-print passage of this subsection.)
Exercise 3.24 You give your ﬁfth grade class a problem:
A faucet ﬁlls a bucket with water in 30 seconds, and the capacity
of the bucket is 12 gallons. How long would it take the same
faucet to ﬁll a vat with a capacity of 66 gallons?
How would you explain to your class how to do this problem?
Exercise 3.25 Consider the following two problems: (a) If you try to
put 234 gallons of liquid into 9 vats, with an equal amount in each vat, how
much liquid is in each vat? (b) If you try to pour 234 gallons of liquid into
buckets each with a capacity of 9 gallons, what is the minimum number of
such buckets you need in order to hold these 234 gallons? Get the answer
to both, and explain in each case whether you are using the partitive or the
measurement interpretation of division.
3.5 The long division algorithm
Suppose we have to do the division problem 7864 ÷19. Up to this point, the
only way we can do it is to look at all the multiples of 19 until we get a whole
number q so that q 19 ≤ 7864 but (q +1)19 > 7864. Of course we could
painstakingly go through all the multiples one by one until we hit one with the
above property, but that would be dull bookkeeping rather than mathemat-
ics. Let us do better. First of all, we can ignore small multiples like 1019 or
even 10019 because all we care about is getting a multiple of 19 that is close
to 7864. Let us make an estimate: the 100th multiple of 19 is 1900, so the
400th multiple is 7600, which is close to 7864. Add ten more of 19 and we get:
7600 + 190 = 7790, which is even closer to 7864. A little experimentation
shows that 7790+319 = 7847 < 7864 and 7790+(194) = 7866 > 7864.
Because 7790+(319) = (41019)+(319) = 41319 by the distributive
law, we know 413 is the quotient. This method of ﬁnding the quotient is
3 The Standard Algorithms 80
clearly superior to the monotonous checking of all the multiples of 19 and
deserves to be made more systematic. With a little more work, this line of
thinking would lead us to the long division algorithm, which is a beautiful
and sophisticated method of ﬁnding the quotient and the remainder of a
division-with-remainder.
Given two whole numbers a, b, with b > 0, our goal is to ﬁnd an eﬃcient
algorithm that produces the quotient and remainder of a ÷b. According to
the second form of the division-with-remainder, this is the same as ﬁnding a
q and an r so that
a = qb + r and 0 ≤ r < b
(See (35).) Let us illustrate the algorithm we are after, long division algo-
rithm, by something relatively simple, say a = 586 and b = 3. Without
further ado, here is the usual schematic presentation of this algorithm for
586 ÷3:
1 9 5
3 ) 5 8 6
3
2 8 6
2 7
1 6
1 5
1
(37)
(Note that for reason of clarity of exposition, we bring the 6 down at each
step of the long division in (37).) We know the conclusion we are supposed
to draw from this: the mechanism described in (37) produces the quotient
195 and remainder 1. In other words,
586 = ( 195 3) + 1 (38)
The question is why ? Because this question can be easily misunderstood,
let us explain it further. The question here is not why (38) is correct; the
correctness of (38) is easily checked, after all, by verifying that 195 3 =
585 so that adding 1 to it produces 586. The question is rather why the
particular procedure adopted in (37), seemingly unrelated to multiplication
or division in the usual sense we understand it, should produce the correct
answer of 195 and 1 in (38). The failure to directly address this question
3 The Standard Algorithms 81
in school mathematics and pre-service professional development materials is
what makes the long division algorithm so notorious. There is nothing wrong
with the algorithm; it has already been stated above that it is one of the most
beautiful pieces of elementary mathematics. There is plenty that is wrong
with the way elementary mathematics is taught, however. Let us proceed
to make amends by explaining the underlying reason why (37) leads to (38)
simply and correctly.
First, we make a general comment about the long division algoithm n
order to clarify our subsequent discussion. You can see from (37) that, in
arriving at the purported quotient 195, the division-with-remainder is used
three times: 5 ÷ 3, 28 ÷ 3, and 16 ÷ 3. Now you may ﬁnd the following
fact puzzling: if we are trying to ﬁnd an algorithm which is more eﬃcient
than the division-with-remainder itself in order to get at the quotient and
the remainder, how can we be using the division-with-remainder itself? Here
is the main point: for “small” numbers, the quotient and remainder of a
division can be easily guessed at (e.g., 5 ÷3, 28 ÷3, and 16 ÷3), so what the
long division algorithm does is to break up the division of a large number
(e.g., 586, although you can easily put up a number as large as you want)
into a sequence of divisions of smaller numbers, and then string the latter
together in an artful way so as to get at the quotient and remainder of the
division of the original large number.
12
Our task is to understand why this
“stringing together” makes sense.
Now we are going to give a preliminary explanation of why the steps in
(37) lead to the correct conclusion (38). It is, we emphasize, only a prelimi-
nary explanation, because we shall subsequently point out in what way it is
unsatisfactory.
The ﬁrst step in (37) looks like 5 ÷ 3, but since 5 stands for 500 in 586,
this particular division is really 500÷3, and the remainder 2 is really 200. So
the ﬁrst step in (33) is actually a restatement of the division-with-remainder
500 = ( 100 3) + 200 . Therefore:
586 = 500 + 86 = ¦([100 3] + 200)¦ + 86
= (100 3) + 286 .
(At this point, we interject a word of caution: there will be many compu-
tations of this type in this subsection, and we have to resist the temptation
12
It is a basic strategy in mathematics to try to break up a complicated task into a
series of simpler tasks.
3 The Standard Algorithms 82
of multiplying out thing like 100 3 above, or 90 3 in the succeeding sen-
tence. The reason is that we want the ﬁnal outcome to be (195 3) + 1
as in (38), and for this reason we want to keep the factor 3 intact all
through these computations.) The second step in (37) is 28 ÷ 3, which as
before is in reality 280 ÷ 3. The division-with-remainder in this case reads:
280 = ( 90 3) + 10 . Thus,
586 = (100 3) + 280 + 6
= (100 3) + ([90 3] + 10) + 6
= (100 3) + (90 3) + 16.
The last step is easy: 16 = ( 5 3) + 1 . So,
586 = (100 3) + (90 3) + (5 3) + 1
= ([100 + 90 + 5] 3) + 1
= ( 195 3) + 1 ,
which is exactly (38). We have of course used the distributive law, and the
reason why this law must play a role is not entirely obvious. It has to do
with (35), where we have pointed out that multiplication and division undoes
each other (at least in the case of no remainder). As we stressed in ¸3.3, the
key reason underlying the multiplication algorithm is the distributive law. It
therefore follows that if division undoes multiplication, and the multiplica-
tive algorithm depends critically on the distributive law, the long division
algorithm must likewise make critical use of the distributive law.
We next give an interpretation of (37) in terms of money. Suppose we
have $586 consisting of
5 hundred-dollar bills
8 ten-dollar bills
6 one-dollar bills.
Although we are trying to ﬁnd out how many 3’s there are in 586, we can turn
the original problem around: suppose there are n 3’s in 586 (with 0 or 1 or
2 left over), then we may also interpret n as the number of dollars in a stack
when 3n dollars are divided equally into 3 stacks. To this end, we begin the
process of creating these 3 stacks by ﬁrst distributing the 5 hundred-dollar
bills equally into these 3 stacks. In each stack we put in 1 hundred-dollar bill,
and there are 2 left over. This corresponds to the ﬁrst step of (37). Next, we
3 The Standard Algorithms 83
convert the 2 hundred-dollar bills into 20 ten-dollar bills, so that (together
with the original 8 ten-dollar bills already there) we now have 28 ten-dollar
bills. These can be distributed into these three stacks equally with 9 in each
stack and 1 left over. This corresponds to the second step of (37). Finally,
we convert the 1 ten-dollar bill into 10 one-dollar bills, and we now have 16
one-dollar bills. Again, we can distribute them equally into the three stacks
with 5 one-dollar bills in each, and 1 is left over. Altogether then, the original
stack of $586 has been divided into three equal stacks each consisting of 1
hundred-dollar bill, 9 ten-dollar bills, and 5 one-dollar bills, with 1 one-dollar
bill left over. This is exactly what (38) says.
Observe that the preceding interpretation of (37) in terms of money is
very similar in spirit to the intuitive approach to ﬁnding a quotient described
at the beginning of this sub-section. There are some people in mathemat-
ics education who consider the use of money to interpret the long divison
algorithm as the height of conceptual understanding. Readers of this mono-
graph cannot fail to realize, however, that the use of money is only an aid
to understanding and is not to be confused with genuine mathematical un-
derstanding itself. We will in fact take up the mathematical explanation
of long division next.
To make sure that the basic facts of the long division algorithm and the
preliminary explanation are understood, we do another example: 1215 ÷35:
3 4
3 5 ) 1 2 1 5
1 0 5
1 6 5
1 4 0
2 5
(39)
Here is an abbreviated explanation:
1215 = 1210 + 5
= ¦(30 35) + 160¦ + 5
= (30 35) + 165
= (30 35) + ([4 35] + 25)
= ((30 + 4) 35) + 25
= ( 34 35) + 25
For an in-depth understanding of the long division algorithm, the pre-
3 The Standard Algorithms 84
ceding analysis falls short in two respects. First, it lacks simplicity. It does
not lead from a clear description of the algorithm straight to the desired
conclusion about quotient and remainder; in fact, a clear description of the
algorithm was never given. Second, the explanation does not clearly expose
the role played by the sequence of remainders (i.e., the numbers 2, 1, 1 in (37)
and the numbers 16 and 25 in (39)) which are critical to the understanding
of the conversion of fractions to decimals in ¸4 of Chapter 4. We now give a
mathematical explanation that is free of these defects.
Let us revisit (37). As we have emphasized throughout our discussion of
algorithms, every one of the standard algorithms gains eﬃciency and sim-
plicity by ignoring place value and by performing the operations one digit
at a time, mechanically. The long division algorithm is the most remarkable
embodiment of these features among the algorithms we have studied. Look
at the “dividend” 586 (i.e., the number to be divided), and we shall describe
the long division algorithm precisely in this special case. The idea will be
seen to be prefectly general. We repeat: the algorithm will go through each
digit of 586, one at a time, with absolutely no thought given to “breaking
the dividend into parts” (as is taught in the schools).
There will be a sequence of steps, each of which performs a division-with-
remainder The divisor will always be the original divisor (in this case
it is 3). So the only thing that needs to be speciﬁed in each step is the
dividend. We start from the left, for a change, and the dividend of the ﬁrst
step is the ﬁrst digit of the original dividend (in this case it is the digit 5 of
586). More formally:
Step 1: perform the division-with-remainder, using as dividend
the leftmost digit of the original dividend.
So the ﬁrst division is 5 ÷3. The division-with-remainder gives
5 = ( 1 3) + 2
The next (second) step is the crucial one, because the algorithm will be
repeating this step ever after. The description of the dividend in the second
step is this:
Step 2: Multiply the remainder of the preceding step by 10 and
add to it the next digit (to the right) in the original dividend.
3 The Standard Algorithms 85
In the present situation, the remainder from the ﬁrst step is 2, and the
next digit of the original dividend 586 is 8. So the number in question is
(2 10) + 8 = 28. Now divide 28 by the same divisor 3:
28 = ( 9 3) + 1
We are now on automatic pilot:
Step 3: repeat step 2.
13
With this in mind, the next digit in the original dividend 586 is 6, so the
dividend of the next step is 1 10 +6 = 16. Thus the third step of the long
division algorithm is:
16 = ( 5 3) + 1
Now (37) is entirely encoded in the following three (simple) division-with-
remainders:
5 = ( 1 3) + 2
28 = ( 9 3) + 1
16 = ( 5 3) + 1
(40)
You could not possibly fail to observe that the quotient 195 is clearly dis-
played in (40) — read vertically down the ﬁrst digits of the right sides —
as well as the remainder 1 (the last term of the last equation). Though of
slightly less interest, you can also read oﬀ the original dividend by going
down the left sides of these equations and pick out the last digit of each
number (in this case, you get 586). If you are baﬄed by these equations, let
us hasten to point out that there is no mystery to them at all. You can, for
example, relate them to the naive interpretation of (37) in terms of money in
the following way. The ﬁrst equation is a restatement of the splitting of the
5 hundred-dollar bills into three equal stacks with 2 left over. The second is
the splitting of the 28 ten-dollar bills into three equal stacks with 1 left over,
and the third is the splitting of the 16 one-dollar bills into three equal stacks
with also 1 left over. But of course one must keep in mind that (40) is true
regardless of any such monetary interpretations.
As we have emphasized, an algorithm is a sequence of mechanical proce-
dures. Steps 1–3 explain how to generate these procedures as we go through
the digits of the original dividend one-by-one. In the case of 586 ÷ 3, (40)
13
Recall what was said at the beginning of ¸3, to the eﬀect that each standard algorithm
would break a computation down to computations with single digit numbers.
3 The Standard Algorithms 86
gives the procedures explicitly. We now show how to generate (38) by the
use of the long division algorithm as encoded in (40):
586 = (5 10
2
) + (8 10) + 6
= ¦([3 1] + 2) 10
2
¦ + (8 10) + 6
by the ﬁrst equation of (40). This gives an explicit support to the interpre-
tation about splitting the 5 hundred-dollar bills into three equal stacks with
2 left over. But to continue:
586 = (3 10
2
) + (2 10
2
) + (8 10) + 6
= (10
2
3) + (20 10) + (8 10) + 6
= (10
2
3) + (28 10) + 6
= (10
2
3) + ([3 9] + 1) 10 + 6
by the second equation of (37), The last line corresponds to the splitting of
the 28 ten-dollar bills into tghree equal stacks with 1 left over. Now apply
the last equation of (40) to get:
586 = (10
2
3) + ([9 10] 3) + 16
= (10
2
3) + ([9 10] 3) + (5 3 + 1)
= (10
2
3) + ([9 10] 3) + (5 3) + 1
= ¦(10
2
+ [9 10] + 5) 3¦ + 1
= (195 3) + 1 .
Let us review what we have accomplished. First we have veriﬁed (38)
directly from a clearly stated digit-by-digit description of the long division
algorithm, which is (40). Second, (40) exhibits the sequence of remainders
2, 1, 1 of the long division algorithm (in the case of 586 ÷ 3) which will be
of critical importance later in Chapter 4. Third, there is a point which has
been purposely suppressed thus far in order to get across the main thrust of
the argument as clearly as possible, but which needs to be aired now. It is
the fact that
at each step of the algorithm ((37) or (40)), the quotients 1, 9,
and 5 are always single digit numbers.
Obviously you have taken this for granted all along, because this is never
mentioned in school mathematics. Nevertheless, the simplicity of the long
3 The Standard Algorithms 87
division algorithm depends on “replacing” each digit of the original dividend
by another digit (e.g., 5 by 1, 8 by 9 and 6 by 5; of course on other occasions,
a digit of the dividend could be replaced by 0) so that this “single-digit” fact
should be carefully veriﬁed. This we proceed to do. It is straightforward to
see from (40) — and this is another reason why the explicit description of the
algorithm in Steps 1–3 is critical — that because each remainder is by deﬁni-
tion < 3, it is ≤ 2, so that the next dividend (i.e., the number 28 after the ﬁrst
step, and the number 16 after the second) is ≤ (20 + a single digit number)
according to Step 2. Therefore the dividend at each step of the long division
algorithm is always < 30. But if a number is < 30, then its quotient when
divided by 3 must be a single digit, exactly as claimed. (Cf. the ﬁrst form of
the Division-with-Remainder in ¸3.4). As usual, we note that this reasoning
is valid in general.
We want to drive home the point that the long division algorithm is
strictly a digit-by-digit procedure without regard to place value. To this end
we shall compare (37) with the division 58671 ÷ 3, where the “dividend”
58671 has been chosen on purpose to have the three digits 5, 8, and 6 at the
beginning. The usual schematic display of the long division is as follows:
1 9 5 5 7
3 ) 5 8 6 7 1
3
2 8
2 7
1 6
1 5
1 7
1 5
2 1
2 1
0
The precise description of the algorithm in accordance with Steps 1–3 is then:
3 The Standard Algorithms 88
5 = ( 1 3) + 2
28 = ( 9 3) + 1
16 = ( 5 3) + 1
17 = ( 5 3) + 2
21 = ( 7 3) + 0
(41)
Now compare the ﬁrst three steps of both (40) and (41). They are iden-
tical. For deﬁniteness of discussion, let us concentrate on the third step in
both. The 6 in 586 is in the ones digit, which is very diﬀerent from the 6 in
58671, which is in the hundreds digit. In other words, the third step in (41)
is actually 1600 ÷3 and the corresponding division algorithm is then
1600 = (500 3)100
if place value is taken into consideration. The point we wish to emphasize
is that, as far as the long division algorithm itself is concerned, place value
is irrelevant. Needless to say, the explanation of why (41) leads to 58671 =
3 19557 — along the line of the argument leading from (40) to (38) —
is squarely based on (41) and nothing else. Contrast this with the ﬁrst
explanation given of (37), which is laden with place value interpretations.
This is the point we made earlier about the lack of simplicity in the latter
explanation.
As a reminder: observe once again that in (41), the dividend 58671 can be
read oﬀ by going down the last digits of the left sides, and the quotient 19557
can be read oﬀ by going down the ﬁrst digits of the right sides. Moreover,
the remainder (0) appears in the last equation.
Let us give an example of a long division where the divisor has more than
one digit, carried out in accordance with Steps 1–3, in order to illustrate
more clearly why it is unnecessary to worry about “breaking the dividend
into parts”. Here then is 11546 ÷19 :
3 The Standard Algorithms 89
0 0 6 0 7
1 9 ) 1 1 5 4 6
0
1 1
0
1 1 5
1 1 4
1 4
0
1 4 6
1 3 3
1 3
Now the digit-by-digit sequence of steps are the following:
1 = ( 0 19) + 1
11 = ( 0 19) + 11
115 = ( 6 19) + 1
14 = ( 0 19) + 14
146 = ( 7 19) + 13
(42)
Now we use (42) to verify that 11546 = 19 607 + 13 . Note that only
the third and ﬁfth equations carry real information, so only those will be
used in the following derivation :
11546 = 1 10
4
+ 1 10
3
+ 5 10
2
+ 4 10 + 6
= (115 10
2
) + (4 10) + 6
= (10
2
115) + (10 4) + 6
= (10
2
([19 6] + 1)) + (4 10) + 6
= (6 10
2
) 19) + (10
2
+ [4 10]) + 6
= (6 10
2
) 19) + 146
= (6 10
2
) 19) + (7 19) + 13
= (([6 10
2
] + 7) 19) + 13
= (607 19) + 13
Hence the quotient of 11546 ÷19 is 607, and the remainder is 13.
Exercise 3.26 Explain to a sixth grader why the long division algorithm
for 642 ÷4 is correct.
4 The Number Line and the Four Operations Revisited 90
Exercise 3.27 Compute 10192 ÷ 8 using the long division algorithm.
Then write out the procedural description of this long division along the line
of (40), and use it to explain why your result is correct, i.e., if the quotient
and remainder of your original long division are respectively q and r, use
your procedural description to show directly that 10192 = (q 8) + r.
Exercise 3.28 Do the same for 21850 ÷43. Be sure you write down every
step of the procedural description (as in (42), for example).
Exercise 3.29 Use (41) to derive the fact that 58671 = 3 19557 . (In
other words, we know you can multiply 3 19557 to get 19557, but we’d
prefer that you learn how to use a sequence of division-with-remainders such
as (41) to explain the long division algorithm.)
Exercise 3.30 Do the long division of 50009÷67 to ﬁnd the quotient and
remainder, describe the algorithm as a sequence of division-with-remainders
in accordance with Steps 1–3, and use these to show why your quotient and
remainder are correct.
4 The Number Line and the Four Operations Revisited
We introduced the number line in ¸2 as an “inﬁnite ruler” with the whole
numbers identiﬁed with a set of equally spaced points (often referred to as
“markers”) to the right of a point designated as 0. A whole number n is also
identiﬁed with the length of the line segment [0, n] from 0 to n. Until the end
of Chapter 4, we shall be concerned exclusively with the part of the number
line to the right of 0.
In the following, we shall always refer to the number line, and this ter-
minology has to be understood in the following sense. The positions of the
whole numbers depend completely on the choice of 0 and 1. Once these two
numbers have been ﬁxed, the positions of the other whole numbers are like-
wise ﬁxed. The segment [0, 1] is called the unit segment, and the number 1
is sometimes referred to as the unit. In each discussion,
we always assume that a unit segment has been chosen on the
given straight line so that the whole numbers are ﬁxed on the
line.
It is in this sense that the number line is ﬁxed. There will be occasions when
we see ﬁt to change the unit, in which case there will be a diﬀerent number
4 The Number Line and the Four Operations Revisited 91
line to deal with. Such an occasion will come up soon enough.
The four arithmetic operations have also been interpreted geometrically.
For addition, we have that for any two whole numbers m and n,
m + n = the length of the segment obtained by
concatenating the segments [0, m] and [0, n]
(43)
Geometrically, we have:

[0,m]

[0,n]
m + n
' E
This way of adding numbers is exactly the principle underlying the slide
rules of yesteryear. (For those who do not know what a “slide rule” is,
perhaps one can describe it as the stone-age version of the calculator.) In
any case, one can do many activities with (43) until this geometric way of
adding numbers become second nature.
Subtraction is next. If m, n are whole numbers and m < n, then we saw
in ¸3.2 that
n −m = the length of the segment obtained when
a segment of length m is removed from one
end of a segment of length n
(44)
In picture:
' E
m
' E
n
n −m
' E
We wish to go into some of the ﬁne points of the addition and subtraction
of whole numbers, especially with respect to the number line. Suppose we
have two number lines, as indicated below.
4 The Number Line and the Four Operations Revisited 92
0 1 2
0 1 2 3 4
What do you say to a student if she tells you that she gets 1 + 2 = 2 in the
following way: She takes [0, 1] from the lower number line and [0, 2] from the
upper number line, and concatenate them as shown:
' E
1 + 2

[0,1]

[0,2]
According to the lower number line, the resulting segment has length 2, and
this is how she gets 2 for her answer. The question is: what is wrong?
In order to explain to her what the mistake is, you would have to recall
for her the fact that all geometric representations of operations on whole
numbers, including (43) and (44), are done on one number line, and there-
fore are done with respect to a ﬁxed unit segment. So what she did wrong
was not to realize that she had changed her unit segment in going from the
upper number line to the lower number line, and subsequently got the two
unit segments mixed up.
This brings up a fundamental issue in the arithmetic operations on whole
numbers concerning
the importance of having the same unit as a ﬁxed reference.
Consider for example the following equations:
9 −2 = 1
8 + 16 = 2
19 + 17 = 3
(45)
Although every equation in (45) is wrong according to the arithmetic of whole
numbers as we know it, it is not as absurd as it appears. What (45) wants
to say is the following:
4 The Number Line and the Four Operations Revisited 93
9 days − 2 days = 1 week
8 months + 16 months = 2 years
19 eggs + 17 eggs = 3 dozen eggs
The point of (45) is to underline the implicit or explicit role played by
the unit in any addition or subtraction of numbers. The concept of a whole
number is an abstract one: for example, the equation 2 + 3 = 5 could mean
any of the following among numerous other possibilities:
2 apples and 3 apples are the same as 5 apples
2 cups of coﬀee and 3 cups of coﬀee are the same as 5 cups of
coﬀee
2 square inches and 3 square inches are the same as 5 square
inches
Whatever the interpretation of the abstract operations, each addition or
subtraction must refer to the same unit. Thus, the ﬁrst interpretation of
2 + 3 = 5 is based on taking 1 to be “one apple”, the second on “one cup
of coﬀee”, and the third on “one square inch”. However, it can never be
interpreted to mean
2 apples and 3 cups of coﬀee are the same as 5 square inches
It is for the same reason — changing the unit in the middle of addition —
that the student’s reasoning of 1 + 2 = 2 is wrong.
In a mathematical context, what we are saying is this. Suppose a ring R
is isomorphic to the integers Z, and suppose under the isomorphism ¯ n ∈ R
corresponds to n ∈ Z. Then although both
¯
2 +
¯
5 and 2 + 5 make perfect
sense, we cannot perform the addition
¯
2 + 5 or 2 +
¯
5.
This discussion of units assumes special importance when we discuss
length and area. In this context, we come to understand (43) as the in-
terpretation of 1 as a ﬁxed unit length, so that each addition is nothing but
combining unit lengths of segments. Now suppose we decide that 1 is the
area of a ﬁxed square. Recall the convention that once we decide on a ﬁxed
segment as the unit segment, then the area of the unit square (= the square
with each side of length equal to that of the unit segment) deﬁnes the unit
area. So the number “1” will henceforth refer to this unit area. Then the
number 3 is no longer a concatenation of three unit segments but rather the
4 The Number Line and the Four Operations Revisited 94
total combined area of three unit squares. For convenience, we shall agree to
interpret a whole number n in this context as the area of a rectangle whose
width is a unit segment and whose length is the concatenation of n unit
segments. For instance, 5 would be the area of the following rectangle:
If there is no fear of confusion, we would simply say this is a rectangle
of width 1 and length 5. Again, in this particular context, 2 + 3 would be a
concatenation of the following two rectangles rather than a concatenation of
the segments [0, 2] and [0, 3] as stated in (43):
(46)
What makes this discussion particularly relevant is that we have inter-
preted the multiplication of whole numbers in ¸2 as area, e.g., 2 3 is the
area of the rectangle:
(47)
Therefore 2 3 = 6 means that the area of the the rectangle of (47) is the
same as the area of
because the latter is exactly what 6 stands for in this context. (Or, to remind
you of the meaning of the equal sign “=” as explained at the beginning of ¸2,
the equality 2 3 = 6 means if we count the number of unit squares on the
left and count the number of unit squares on the right, the two numbers are
4 The Number Line and the Four Operations Revisited 95
the same.) To drive home this point, we give one possible interpretation of
5+(23) = 11: it means that the area obtained by combining the rectangle
of (46) and the rectangle of (47) is the same as the area of the rectangle
whose width is 1 and whose length is 11. We shall be discussing the area
interpretation of numbers extensively in Chapter 2.
Two extra comments would help to clarify the circle of ideas in connec-
tion with the representation of 1 as the area of the unit square.
The ﬁrst one is that declaring the area of the unit square (recall: this
is the square whose side has length 1) to be 1 is nothing more than a
convention. In fact, we could declare its area to be any number, say, 2.
What could go wrong then? The area formula for the rectangle would be
messed up, as follows. Look at the area of the rectangle with width 5 and
length 7, for example. This rectangle is paved (tiled) by 5 7 unit squares:
Therefore the area of the rectangle is 2 (5 7), and not (5 7) (which
would be the case if each unit square has area equal to 1). More generally,
if a rectangle has width m and length n, then its area would be 2mn intead
of the usual mn. Thus declaring the area of the unit square to be anything
other than 1 would serve no purpose other than messing up otherwise simple
formulas. For this reason, we all agree to set the area of the unit square to
be 1.
A second comment is that the representation of the product of whole
numbers as the area of the corresponding rectangle has a long history be-
hind it. Until the time of Descartes (1596-1650), this was the only way to
understand the multiplication of numbers. In the most inﬂuential mathe-
matics textbook of all time, Euclid’s Elements (circa 300 B.C.E.), there was
never any mention of multiplying two numbers m and n. Each time Euclid
wanted to express that idea, he would say: “the rectangle contained by the
line m and the line n” (translation: “the rectangle” in Euclid means “the
area of the rectangle”, “contained by” means “having for its sides”, and “the
line m” means “the line segment of length m”). For this same reason, a prod-
uct of three numbers, such as 12 7 9, had to be interpreted as volume
4 The Number Line and the Four Operations Revisited 96
and therefore the product of four or more numbers was almost never consid-
ered until Descartes pointed out that multiplication can also be regarded as
an abstract concept independent of geometry. Nowadays, a (good) college
course on number systems would develop all number concepts in an abstract
setting without reference to geometry. For this monograph, however, the
geometric interpretation of multiplication not only is convenient for our pur-
pose, but has the added advantage in that it is suﬃciently similar to the
common manipulative of Base Ten Blocks to make a beginner feel at ease.
The purely algebraic approach to multiplication will be discussed in ¸7.2 of
Chapter 2 and also ¸3 of Chapter 5.
At this point, we are in a position to give another interpretation of the
multiplication of whole numbers, one that was mentioned in ¸2. The product
34, for example, can be interpreted as the number 3 on a number line whose
unit 1 is taken to be (the magnitude or size represented by) the number 4;
one can think of 4 as a bag of 4 potatoes, a car full of 4 people, a box of 4
crayons, a bag of 4 marbles, etc. Schematically, in terms of such a choice of
the unit 1, the number 3 is the point on the line represented by the following
3 groups of objects:
• •
• •
• •
• •
• •
• •
In general, if m and n are whole numbers, then mn may be interpreted as
the number m on the number line whose unit 1 is taken to be (the magnitude
or size represented by) the number n. So m in this context is the point on
the line represented by the following m groups of objects:
n objects n objects

n objects

m
For a later need, it would be advantageous to formalize this procedure.
To facilitate the discussion, let us give m and n explicit values, say m = 4
and n = 3, and we shall re-interpret 4 3 as the point 4 on a number line
with a new unit. So we start with a number line:
4 The Number Line and the Four Operations Revisited 97
0 1 2 3 4 5 6 7 8 9 10 11 12
Now introduce new markers on the same line, where the new unit is 3.
To avoid confusion, we shall distinguish the new number markings from the
original one by a bar and place them underneath the line. So 1 is right under
3, 2 is right under 6, etc. In particular, 4 3 is four copies of the new unit
and is therefore 4 in the new number line, and 3m would be just m for any
whole number m.
0 1 2 3 4 5 6 7 8 9 10 11 12
1 2 3 4
This idea of using a new unit to re-interpret the multiplication of numbers
provides an alternative way to understand the multiplication of fractions, as
we shall see in ¸7.3 of Chapter 2.
Finally, division. For whole numbers a, b, with b always assumed to be
nonzero, let us assume for now that a is a multiple of b, say a = qb for some
whole number q. Then
we write a ÷b = q for a = qb. (48)
In intuitive language, multiplication and division undoes each other. (See
(36) and the discussion surrounding it.) Sometimes, we also say multiplica-
tion and division are inverse operations to express this fact.
Activity: Suppose a fourth grader understands 24 ÷ 3 = 8
only in terms of the measurement interpretation and the partitive
interpretation of division. Explain to her why 24 ÷ 3 = 8 is the
same as 24 = 8 3.
This is the place to tie up a loose end mentioned in the discussion of the
division algorithm, namely, why one cannot divide by 0. Suppose, division
by zero makes sense for a particular nonzero whole number n, say n÷0 = 3.
If one gives a little thought to what m ÷ n = k could mean regardless of
4 The Number Line and the Four Operations Revisited 98
what m, n, k may be, one would likely conclude (with (48) as guide) that
it means m = nk. In other words, one would require that (48) makes sense
for all m, n, and k. Such being the case, n ÷ 0 = 3 would be the same as
saying n = 0 3 = 0, which contradicts our assumption that n is nonzero.
If 3 is replaced by any other whole number, the same argument applies. So
n ÷0 cannot be equal to any whole number if n is nonzero, which is another
way of saying it cannot be deﬁned. What about 0 ÷ 0 ? We now run the
preceding argument backwards, i.e., m = nk should mean the same thing as
m÷n = k for all m, n. Thus knowing 0 = 0 1 means 0 ÷0 = 1 . But it
is also true that 0 = 02 , so 0÷0 = 2 . We have therefore shown that the
value of 0 ÷0 is ambiguous; it could be 1, or 2, or in fact any whole number
by the same argument. This shows that 0 ÷ 0 cannot be given a deﬁnite
value, i.e., it is also undeﬁnable. We have therefore shown that division by 0
cannot be deﬁned.
It may be instructive to follow literally the partitive and measurement
interpretations of division to see why n ÷0 cannot be deﬁned for a nonzero
whole number n. Suppose it were deﬁnable. By the partitive interpretation,
n ÷0 would mean the number of objects in a part when n objects are par-
titioned into 0 equal parts (see ¸3.4). Because we cannot partition anything
into 0 equal parts, this has no meaning. Now suppose n÷0 were meaningful
in the measurement sense. Then it is the number of parts when n objects are
partitioned into diﬀerent parts so that each part has exactly 0 objects. But
if each part has no object, the partition of the n objects cannot be done. So
again, this interpretation has no meaning either.
The fact that division undoes multiplication (always understood in the
sense of (36) or (48)) leads to a geometric interpretation of division that has
already been mentioned at the end of ¸3.4. Assuming as always that a is a
multiple of b, then a ÷b is the other side of a rectangle whose area is a and
one of whose sides is equal to b:
a a ÷b
b
(49)
To conclude this discussion of division, it remains to point out that al-
5 What Is a Number? 99
though the restriction that a is a multiple of b imposed here seems too severe,
it will be seen when we come to ¸8 of Chapter 2 that it is in fact no restric-
tion at all once fractions are at our disposal. The equivalence of division and
multiplication as described in (48) will be seen to be the key to the under-
standing of division in general.
Exercise 4.1 Use the idea of introducing a new unit to represent a whole
number c to re-interpret the distributive law in the form of (a+b)c = ac+bc.
Exercise 4.2 If a rectangle has area 98 and one side equals 14, what is
the other side? If the area is 1431 and one side is 27? And if the area is 7797
and one side is 113?
5 What Is a Number?
Thus far, we have never paused to ask what a whole number is. We took
this concept for granted from the beginning, and subsequently make them
correspond to a collection of equally spaced markers (points) on a line to the
right of a point denoted by 0. We are going to change our viewpoint here
and use a set of markers on a line to deﬁne the whole numbers. Before you
ask why we bother, let us do it ﬁrst. So we start afresh by imposing a set of
markers on a line.
Take a straight line and mark oﬀ a point as 0 (zero). Then
ﬁx a segment to the right of 0 and call it the unit segment. Mark
the right endpoint of this segment on the line, thereby generating
the ﬁrst marker. Slide the unit segment to the right until its
left endpoint is at the ﬁrst marker; mark the new position of the
right endpoint of the unit segment, thereby generating the second
marker. Now slide the unit segment to the right again until its left
endpoint rests on the second marker, and mark the right endpoint
of the unit segment in its new position. This generates the third
marker, etc. This generates a sequence of equally spaced markers
to the right of 0.
Notice that up to this point, there is no mention of whole numbers. Now
we will formalize the introduction of whole numbers by adopting the follow-
ing deﬁnition.
5 What Is a Number? 100
Deﬁnition. A whole number is one of the markers on the line, so that,
starting with the initial number 0, the next one (to the right of 0) is 1, the
one after that is 2, etc., and we continue the naming of the markers in the
same way we did the counting of the whole numbers in ¸1. This line with
the whole numbers on it is called the number line. A number is by deﬁnition
any point on the number line.
As far as whole numbers are concerned, what we have deﬁned is at least
consistent with everything we have done up to this point. In this sense, we are
not by any means trying to attempt a retrograde revision of our knowledge
of the whole numbers. Rather, we are saying that, for the work we do from
now on, we shall agree to change our point of view and base our reasoning
with whole numbers on this deﬁnition alone. Thus a whole number is now
something very concrete and explicit: it is among the markers on the number
line which were carefully constructed above. Because we have been using
these markers all along in our work, no revision of anything we have done is
necessary. Whatever we do in the future about whole numbers, however, we
should be able to explain it in terms of these points on the number line.
You must be muttering to yourself at this point and wondering what is
happening here. After all, don’t we know what a whole number is? Let us
deﬁne the number 5, for instance. Note that what is required here is not a
description of our intuitive feelings about “5”, but rather a precise deﬁnition
analogous to the deﬁnition (say) of a triangle as three noncollinear points to-
gether with the line segments joining them. We know “ﬁve ﬁngers”. We also
know “ﬁve chairs”, or “ﬁve apples”, or ﬁve of anything we see or touch be-
cause we can count. But ﬁve itself, without reference to any concrete object?
So you see that it is diﬃcult. Do not be discouraged, because the general
concept of a number, in the sense deﬁned above as a point on the number
line, baﬄed the human race for over two thousand years before it was ﬁnally
pinned down in the late nineteenth century. What we need for elemetary
mathematics is fortunately nothing very sohpisticated, just the whole num-
bers and some other numbers which come out of whole numbers in a rather
simple-minded fashion. Fractions or rational numbers, for instance. In other
words, we will not scrutinize every point on the number line. Only a small
portion of those points. This deﬁnition of numbers is not ideal, but it serves
our pedagogical needs admirably, in the sense that it is accessible and it
lends itself to a reasonable treatment of rational numbers and decimals. See
Chapters 2, 4 and 5.
5 What Is a Number? 101
A precise deﬁnition of whole numbers is not strictly necessary if all we
ever do in mathematics is to stay within the realm of whole numbers. For
example, even if we cannot deﬁne precisely what 5 is, we can communicate
the essence of it by putting up one hand with the ﬁngers outstretched; that
should be enough to communicate any kind of “ﬁveness” needed for con-
ceptual understanding. This is the advantage of whole numbers: each has
(at least in principle) a concrete manifestation such as outstretched ﬁngers
that almost renders abstract considerations about whole numbers unneces-
sary in elementary school. But we cannot stay with whole numbers forever,
because the next topic is fractions. What concrete image can one conjure
in connection with
13
7
or
119
872
? Children need answers to this question in
their quest for knowledge because they need something to anchor the many
concepts related to fractions in the same way a handful of ﬁngers can anchor
any discussion about 5. Amazingly, school mathematics in our country has
contrived to never answer this question. The results are entirely predictable:
when adults abrogate their basic responsibilities, the ﬁrst victims are the
children. The generic non-learning of fractions among children has become
part of our national folklore, so much so that you can ﬁnd references to it in
the comic strips of Peanuts and FoxTrot. We want to change this dismal
scenario by adopting the down-to-earth approach of
giving direct answers to direct questions.
We will deﬁne fractions, decimals, or any concept we ever take up.
Now you would want to know why not just deﬁne fractions abstractly and
leave a nice subject like whole numbers alone. The answer is that in order
for children to understand fractions, fractions cannot be suddenly presented
to them out of the blue. Learning is a gradual process ﬁrmly rooted in prior
experiences. If we can convince them that fractions are nothing more than
an natural extension of the whole numbers, then our chances of success in
teaching them fractions would be immeasurably increased. At the moment,
most (or perhaps all) of the school textbooks and professional development
materials would have you believe that whole numbers are simple, but frac-
tions are a completely diﬀerent breed of animals. Whole numbers are taught
one way, and fractions in a completely diﬀerent way. There is no continuity
from one to the other. The minute you as a teacher or your students buy
5 What Is a Number? 102
into this view of numbers, our mathematics education is already in trouble
because this means you have bought into mathematical misinformation.
From the point view of mathematics, whole numbers are on an equal footing
with fractions. They are part of the same family, the real numbers. In fact,
this is where the number line comes in: we already have the whole numbers
there, and the next step is to single out the fractions on this line. For this
reason, it is not possible to only oﬀer a precise deﬁnition of fractions and
leave the whole numbers unattended. We must begin with a deﬁnition of the
whole numbers that will naturally lead to fractions.
Let us consider the philosophical question of why something as natural as
a whole number should be made into something as cold and formal as “a point
on the number line”. The answer lies in the fact that we are trying to deepen
our understanding of the whole numbers in order to lay the groundwork for
working with fractions. In particular, the geometric interpretations (43)–
(44), (46)–(47) and (49) are part of this groundwork. Now it is in the
nature of human aﬀairs that each time we try to achieve excellence in any
endeavor, doing what is natural is simply not enough. Take running, for
instance. This is about as natural an activity as we are going to get. In
fact, had our ancestors been less good at it, all of them would have been
hunted down by the predators on the African Savannas and we wouldn’t be
here to talk about fractions. Yet, if you talk to an Olympic sprinter, what
you hear from him about running would strike you as extremely unnnatural
if not downright unreal. He would tell you that he calculates exactly how
many strides he takes with each breathe-in, how many he takes with each
breathe-out, exactly how far from the starting block before he can take his
ﬁrst breath, and where the next spot is before he can take another breath
again. Doesn’t this unnnatural and calculated approach to running remind
you of looking at a whole number as a point on the number line?
But let us not lose our perspective. Whatever the Olympic sprinters do
in a race, it is highly unlikely that they think about “how soon before I can
breathe in and how many strides I should take before then” each time they
run to catch a bus. In the same way, you need not ﬁxate on the number
line every time you count oranges in the supermarket. All you need to do
is to understand decimals and fractions and all the rest, and be good teach-
ers. So please rise to the occasion when there is a need to regard a whole
number as a point on the number line and be willing to work with this con-
cept. If you can accept this reality, then you have already won half the battle.
6 Some Comments on Estimation (not yet written) 103
6 Some Comments on Estimation (not yet written)
7 Numbers to Arbitrary Base (not yet written)

2

How to Read This Monograph In order to learn mathematics, you will have to read articles and books in mathematics, such as this monograph. Reading mathematics requires a diﬀerent skill from reading novels or magazines, and I want to say a few words about this diﬀerence. First of all, reading mathematics requires sustained eﬀort and total concentration. It is a slow and painstaking process. This monograph is not a great candidate for bedtime reading unless insomnia is not a problem with you. Some mathematics textbooks, especially those written in the last three decades, do a lot of “padding”, i.e., inserting long passages with little content, and may have instilled the illusion that skipping is a good policy in the reading of mathematics. This monograph, by contrast, says only what needs to be said, so you will have to read every line and try to understand every line. In fact, often you will ﬁnd yourself struggling to understand every word in order to move forward. On occasion, I do some chatting, but more often than not I will be talking about straight mathematics. I have made every eﬀort to supply you with suﬃcient details to follow the reasoning with ease, but I tend not to waste too many words. So you would have to read everything carefully. In the event that I believe something can be safely skipped on ﬁrst reading, I leave it in indented fine-print paragraphs. You may have gotten used to the idea that in a mathematics book you only need to look for soundbites: understand a few procedures and forget the rest. Not so here. This monograph tells a coherent story, but the outline of the plot (the procedures) is already familiar to you. It is the details in the unfolding of the story (the reasoning) that are the focus of attention here. Think of yourself as a detective who has to solve a murder case: you already know going in that someone was killed, so you job is not just to report the murder but to to ﬁnd out who did it, how he did it, and why he did it. It is the details that matter, and they matter a lot. Learning the details of anything is hard work. I want to to tell tell you that most mathematicians also regard learning mathematics as very hard work. It takes time and eﬀort, and it may mean being stuck for a long time trying to understand

3

a particular passage. Nothing good comes easily. It would be futile, not to say impossible, for me to anticipate the kind of diﬃculties each of you may have in reading the monograph. Experience would seem to indicate, however, that most of you will be surprised by the emphasis in this monograph on the importance of definitions. A (very) mistaken belief which unfortunately has gained currency in recent years is that, in the same way that children learn to speak whole sentences without ﬁrst ﬁnding out the precise meaning of individual words, students can also learn mathematics by bluﬃng their way through logical arguments and computations without ﬁnding out the precise meaning of each concept. As a result, it is customary in schools to teach mathematics using mathematical concepts that are only vaguely understood. Such a belief is completely without foundation. Take the concept of a fraction or a decimal, for example. It is almost never clearly deﬁned. Yet children are asked to add, multiply and divide fractions and decimals without knowing what they are or what these operations mean, and textbooks contribute to children’s misery by never deﬁning them either. If we can get away with this kind of mathematics education, — in the sense that children learn all they need to learn without the beneﬁt of clear deﬁnitions — ﬁne. But we cannot, because children are on the whole not learning it. From the standpoint of mathematics, the ﬁrst remedy that should be tried is to explain clearly what these concepts mean, because mathematics by its very nature is a subject where everything is clearly explained. Giving clear deﬁnitions of concepts before putting them to use has the virtue of taking the guesswork out of learning: every step can now be explained, and therefore more easily learned. This is the approach taken here. If you feel uncomfortable with such an approach, can you perhaps suggest an alternative? In any case, it is only a matter of time, and maybe a little practice, before you get used to it. (Smokers also feel extremely uncomfortable at the beginning of their attempt to quit smoking.) You will discover that having clear-cut deﬁnitions is by far the better way to learn and to teach mathematics. At the risk of stating the obvious, I may point out that while

Because mathematics is somehow different. Whatever the justiﬁcation of this kind of misconception. The thinking goes roughly as follows. We should never forget that mathematics is an integral part of human culture. Please keep this in mind as you read this monograph. Why this is worth mentioning is that there is at present a perception that mathematical writing should not be couched in ordinary English.
. and complete sentences are optional. rather than in spite of it. and it imposes on us the same obligations of normal human communication as any other endeavor. for instance. we would do well to ﬁrst change this misconception about mathematical writing amongst ourselves. it requires a different kind of writing: fewer words and more symbols. The subject matter requires greater precison of expression than a chat about the private lives of movie stars. to be sure. but this precision is something we try to achieve in the context of normal communication. If we want to change such behavior among students.4
this monograph addresses serious mathematics. its exposition is given in ordinary conversational English (or as conversational as an ESL person can manage it). We must make ourselves understood via the usual channels in the usual manner. Doing mathematics is above all a normal part of human activities. and a random collection of symbols out of context usually passes for an explanation. the end result is there for all to see: students stop using correct gammar and syntax in their homework and exam papers.

namely. — the arithmetic of whole numbers — with the goal of explaining everything along the way. be it elementary or advanced. there is an inherent danger that as you read this material you would put yourself on automatic intellectual pilot and cease to think. 4. with a view towards laying a ﬁrm foundation for the treatment of the main topics of this monograph. I would explicitly ask you to put yourself in the position of a ﬁrst-time learner and to make believe that you are encountering every topic for the ﬁrst time. 1. 3.
. learning this material is so crucial for the understanding of the rest of the monograph that I must ask you to please try your best. your obligation is to keep alive this sense of curiosity in a child. One way to do this is to ask yourself the same question at all times and to ﬁnd out the answers. 2. . you would not only acquire an enhanced appreciation of the marvelous qualities of many things you have always taken for granted. I realize that it is very diﬃcult to do this because it requires a suspension of habits. To try to ﬁnd out why something is true is a very natural human impulse. Nevertheless. Should you have any doubts. but also ﬁnd yourself in a much better position to learn the materials in the later chapters on decimals and fractions. The main emphasis throughout will not be on the well-known procedures such as the long division algorithm — although a precise and correct statement of that algorithm is certainly diﬃcult to ﬁnd in the literature — but on the logical reasoning that underlies these procedures. just observe how often pre-school children raise this simple question with their parents each time they are introduced to something new. this chapter will revisit a very familiar territory. Notice that we include 0 among the whole numbers. . As a teacher. the ﬁrst question you should always ask when confronted with any statement is “Why?”. fractions and decimals. Once you get the hang of it. To this end. To get you out of this counter-productive mode. .5
Chapter Preview This chapter discusses the whole numbers 0. because it is also your obligation to answer this question for your students. Because everything here is familiar to you. at least as far as procedures are concerned. In mathematics.

This chapter discusses only the whole numbers 0. . . 1.D. These examples illustrate a fundamental and fruitful idea of representing numbers. . . 8.e. .. 2.e. . and the discussion of this numeral system will continue in Chapters 4 and 5. The much needed research has not yet been done. the “4” in the second place from the right stands for 40.1 It became the universal numeral system in the West circa 1600.
This is most likely a misnomer. . As is well-known. 1. two thousand. Thus the “5” in 125 represents a completely diﬀerent order of magnitude from the “5” in 2541. themselves. no matter how large. and by the use of their place in the number symbol to represent diﬀerent magnitudes (sizes). the so-called Hindu-Arabic numeral system. In fact. i. 2. being in the fourth place (position) from the right. as mentioned above. 1. 2. Moreover. is identical to the Hindu-Arabic system except for the ten symbols 0. The term place value means that the value (magnitude. . Similarly. 2. the digit “5” being in the third place from the right stands not for 5.. . the ten symbols 0. The digit “2” in 2541.C.e. .. i. The Chinese had a decimal numeral system since the ﬁrst available record of writing dating back to at least 1000 B. 5 and 500 respectively. . namely.1 Place Value
6
1 Place Value
One cannot understand the arithmetic of whole numbers without a basic understanding of our numeral system. Similarly. We are here mainly concerned with the fact that a symbol such as 2 in the number 2541 stands not for 2 but 2000. . two thousand ﬁve hundred and forty-one. it may be diﬃcult to separate what is Indian and what is Chinese in the Hindu numeral system. and the rod numeral system (also called the counting board numeral system) which has been ﬁrmly in place no later than 200 A. by the use of only ten symbols 0. and so on. and “1” being in the right-most place means just 1. 64738 denotes 60000 + 4000 + 700 + 30 + 8 and 6001 denotes 6000 + 1. ﬁve hundred. 1. 2541 means 2000 + 500 + 40 + 1 . . Because of the long history of contact between the Indians and the Chinese. size) of each digit depends on its place in the numeral symbol.
1
. but for 500. . 9. 3. i. 9 are called digits. .. negative numbers and decimal fractions (see Chapter 4) have been part of the rod numeral system from the beginning. stands not for 2 but 2000.

Observe then that the three “X’s” are in three diﬀerent places. and only the third stands for 1 itself. Just 10. there is a partial place value at work. 9 — instead of twenty or sixty. Contrast this with the numeral 111 in our numeral system: the ﬁrst 1 on the left stands for 100. 1000. used a numeral system that used place value only partially. . People seem to have no problem with understanding this concept.
. let us use it to illustrate how complications arise from trying to cope with large numbers when place value is not systematically applied. the fact that only ten symbols are used is easy to overlook due to constant usage. No other numeral system of the world (except the rod numeral system of China. yet each and every one of them stands for 10. let us look at a diﬀerent numeral system for comparison: in Roman numerals.e. Like place value.1 Place Value
7
In case the idea of place value has become too commonplace to strike you as noteworthy. 2. but that would be too pedantic. We now ﬁll in this gap. In order to write a million. therefore. The Babylonians in the B. to be exact) — to generate all the numbers. 125 = 100 + 20 + 5). . era. Similarly. but they all stand for 1. We could have pretended that you didn’t know what it means to add whole numbers and give a precise deﬁnition. which denotes a thousand. which is a thousand thousand. It should be pointed out.. no matter how large. for instance. For instance “VI” is 6 while ”IV” is 4. not 100 or 1000. multiplication. say — are used to denote any number. The symbol with the largest numerical value in the Roman system is M . i. So far we have not brought out the signiﬁcance of the fact that only ten symbols 0. .2 the number 33 is represented by XXXIII. You see the diﬀerence. the second stands for 10. 1. that the great virtue of the Hindu-Arabic system lies precisely in the systematic and combined application of both ideas — place value and a ﬁxed small number of symbols (ten. But we will carefully and precisely deﬁne the other three arithmetic operations in this and the the next two sections. the three “I’s” occupy diﬀerent places too. 8. period. We have used the concept of addition to explain place value (e. see footnote 1) has ever attained the same degree of symbolic economy. Since we have already mentioned the Roman numerals. subtraction. and division.g. and the symbolic representation of some numbers became unwieldy. one would have to write
2 It is well to point out that even in the Roman numeral system..C. .

. .1 Place Value
8
M M M . 8. . 1. After 0. Even such a desperate eﬀort cannot save this numeral system from certain disaster.e. 08. . Would you want to waste your time learning how to navigate in such a system? To truly understand place value. it is natural to increase the latter from 1 to 2 and start the counting in the ones place all over again with 0. 02. 1. . Continuing this way. 99. 12. For instance.3 From this point of view. and start the counting in the ones place all over again with 0. the so-called tens place. what about writing 60. . 19. because after having used up all the digits in the ones place with the tens place occupied by 1. The Romans were spared this drudgery apparently because they never had to deal with a large number of this size. more symbols would yet be introduced. Thus X would denote not 10 but 10000. Thus we change the 0 of 09 to 1. . 9. the next number after 9 is naturally 10. someone would have tried to introduce symbols for “ten thousand” and “hundred thousand” to simplify the writing of 388999. we have used up all possible symbols by allowing ourselves only one place (position). . M M (a thousand times). At
There is a further discussion of this issue of having 0’s to the left of a number at the end of this section. (Keep in mind that this is no more than a convention. 8. 11.
3
.037? Presumably. the simplest way to write 388999 is CCCLXXXV IIICM XCIX. 22. 21. the only option available is to put the same synbols in an additional place. and M would denote a million. Thus the next numbers are 20. however. the socalled ones place. i. . 18. onward in order to see how the whole numbers develop in the Hindu-Arabic numeral system. etc. we must review the process of counting from 0. .. The same outlook then guides us to write the number after 19 as 20. The numbers after 09 are then 10. . . 2. 1. which by convention is to the left of the ones place. .845.) In our minds. A latter day ad hoc convention to improve on the Roman system is to add a bar above each symbol to increase its value a thousand times. in other words the single-digit numbers may be thought of as two-digit ones with 0 in the tens place. we replace 0 by its successor 1.279. but if so. It is conceivable that had Roman numerals been adopted as the universal numeral system in the modern era. we get to 97. To generate the next set of numbers without adding more symbols. . . . 9 as 00. since 0 has already been used in the tens place. . . . 09. 2. 01. 98. because it would still be too clumsy. . we may think of 0. .

In general. 0. we go from 0 to 100. . to be followed by 101. 0. . then a precedes b. 9 00 . 0 (n zeros) from 0. we have used up all ten digits in both the tens place and the ones place. then we have 00. We pause to note that. . and without introducing more symbols. . we get to 1000 . the so-called thousands place (fourth place from the right). . the so-called hundreds place (the third place from the right). tens. . 112. 100. . . 0 (n zeros) in steps of 100 . 113. 0. To proceed further. 1 000 . . . again we have used up all ten digits in three places so that the next number will have to make use of a fourth place. . 9 (n times for any nonzero whole number n). . we make the following observation: if we count from 0 to 100 . 2 00 . . . It will have four digits and it has to be 1000 because 999 can be thought of as 0999 and we naturally increase the 0 of 0999 to 1 and start counting all over again in the ones. . in ten steps of 100 . Thinking of 99 again as 099. (ii) given two n digit numbers a and b. . the next number must be 100 . and the next is 110. . (i) an n digit number precedes any number with more than n digits. 201. .1 Place Value
9
this point. 80. etc.. we will have to make use of another place to the left of the tens place. if the n-th digit (from the right) of a precedes the n-th digit of b. the same consideration then dictates that the next number is 100. 20. . After that come 200. for exactly the same reason. . . . 0 (n zeros). between the numbers 0 and 100. . 90. 0’s is 1 00 . 202. . . . By the time we reach 999. . whenever we reach the number 99 . and (iii) the sum of ten 1 00 . . then we have 0 00 . . 10. . . 0 (n − 1 zeros). Thus in ten steps of 10’s. 199. if we skip count by 10’s. Thus we come to 109. 0 . 102. 198. 0. As before.
n−1 n
In view of the previous comments about adding 0’s in front of a number.
n−1 n−1 n−1 n−1 n
In other words. . followed in succession by 111. . . . and hundreds places. . etc. . Thus we can make three observations about the way counting is done in the Hindu-Arabic numeral system: for any nonzero whole number n. 1 00 . 0. 0 (n − 1 zeros) for any nonzero whole number n. . we should add the following clariﬁcation: a number is said to be an n digit
.

10
This is the right place to review and make precise the common notion of “bigger than”. 0050000 is a 5-digit number. then a < b. counting from the right.
Note that one sometimes says greater than in place of bigger than. a to be smaller than b) if. and less than in place of smaller than. 65739 prcedes 70001. a comes before b. As illustrations of (i)–(iii): 987 precedes 1123. If we want to allow for the possibility that b is bigger than or equal to a. 100002 > 99817. and 1234 is a 4-digit number.
Thus 13 ≤ 13 and 7 ≥ 7. for two whole numbers a and b. and 10000 + 10000 + · · · + 10000 = 100000. the last nonzero digit is in the n-th place. we deﬁne b to be bigger than a (or what is the same. then we write: a≤b or b ≥ a. In particular. 872 < 1304. b are two whole numbers with n digits and the n-th digit of a is smaller than the n-th digit of b. and (v) if a. but 7 < 13 and 9356 < 11121. etc. this means: 30000000 + 1000000 + 500000 + 30000 + 6000. As we know. For example. In symbols: a<b or b > a. then a < b. Formally. n > 0.
. in the method of counting described above. b are whole numbers and b has more digits than a.1 Place Value
10
number if. 803429 < 911104. We next turn our attention to the phenomenon of “too many zeros”. For example. it is always the case that if n is a nonzero whole number. then It follows from observations (i) and (ii) that (iv) if a. etc. Consider a moderate-size number such as the number of seconds in a 365day year: 31536000.

1 Place Value

11

I hope you are already weary of reading and trying to keep track of so many zeros. It could be much worse, of course. For example, the age of the universe has now been established (by observations of the Hubble telescope) to be approximately 14000000000 (9 zeros) years. Back in the third century B.C., Archimedes estimated that the number of grains that can be packed into a ball the size of the then-known universe was 1, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000 (63 zeros). Clearly, we must devise a shorthand notation to deal with these zeros. (In case you haven’t noticed, symbolic notation has always arisen from the need for better human communication. The Hindu-Arabic numeral system — as you have seen — is a good example.) We need to introduce a new notation and to review an old one. First the new notation: we write 101 for 10, 102 for 100, 103 for 1000, 104 for 10000, 105 for 100000, and in general, write 10n for 1 00 . . . 0
n

(1)

where n is any whole number > 0. For notation consistency, we also write 100 for 1.

(You may think of 100 as “zero number of 0’s”.) The number n in 10n is called the power or exponent of 10n , and 10n is read as “10 to the n-th power”. Next, we recall the deﬁnition of multiplication among whole numbers as a shorthand notation for repeated addition. In other words, 3×5 by deﬁnition 5 + 5 + 5 , and in general, if m, k are whole numbers, the deﬁnition of mk (the accepted abbreviation of m × k) is:   0 if m = 0 mk = (2) k + k + · · · + k if m = 0 
m

Sometimes we refer to mk as the product of m and k, and call m and k the factors of mk. We call attention to the fact that mk, the multiplication of k by m, is a shorthand notation for adding k to itself m times, no more and no less. Please be sure to impress this fact on your students. (See Exercise 1.8.)

1 Place Value

12

Activity: Consider the following introduction to multiplication taken from a third grade textbook (the text has the goal of making sure that at the end of the third grade, students know the multiplication table of numbers up to 10): Look at the 3 strips of stickers shown on the right (there is a picture of three strips of stickers). There are 5 stickers on each strip. How can you ﬁnd the number of stickers there are in all? You can ﬁnd the total number in different ways. You can write an addition sentence. 5 + 5 + 5 = 15 Think: 3 groups of 5 = 15. Answer: 15 stickers. Do you think this is an ideal way to convey to third graders what multiplication means? The deﬁnition of multiplication in (2) lends itself to an easy pictorial representation. For example, 3 × 5 , which is 5 + 5 + 5 , can be represented by three rows of ﬁve dots: • • • • • • • • • • • • • • •
m

There should be no mistaking the fact that the deﬁnition in (2) means exactly that mk is adding k to itself m times, and NOT adding m to itself k times. We are taking nothing about multiplication for granted, so that if we wish to say mk actually also equals adding m to itself k times, we would have to explain why. This will be done later in §2, but for now we don’t need this distraction. We now put the new information to use: if n is a nonzero whole number, then 2 × 10n = 1 00 . . . 0 +1 00 . . . 0 = 2 00 . . . 0
n n n

We can now revisit 31536000 and rewrite it as 31536000 = (3 × 107 ) + (1 × 106 ) + (5 × 105 ) + (3 × 104 ) + (6 × 103 ). (We recall the convention concerning parentheses: do the computations within the parentheses ﬁrst.) Similarly, the age of the universe is approximately 14, 000, 000, 000 = 10, 000, 000, 000 + 4, 000, 000, 000 = (1 × 1010 ) + (4 × 109 ), and Archimedes’ number of grains of sand is simply 1063 . In general, a whole number such as 830159 can now be written as 830159 = (8 × 105 ) + (3 × 104 ) + (1 × 102 ) + (5 × 101 ) + (9 × 100 ) . Such an expression is sometimes referred to as the expanded form of a number, in this case, 830159. Another example is 2070040 = (2 × 106 ) + (7 × 104 ) + (4 × 101 ). One advantage of writing a number in its expanded form is that it instantly reveals the true value of a given digit in the number. For example, the 3

. 0 . such as when we discuss the complete expanded form of a number later on in §3 of Chapter 4. The notation also gives a clear and precise location (place) of a digit in the number symbol: the exponent.
n
whereas we have deﬁned 10n in (1) as 10n = 1 00 .1 Place Value
14
of 830159 is given on the right as 3 × 104 . we will use 5 × 101 and 9×100 . It remains for us to tie up some loose ends in the foregoing discussion.
. On the other hand. we have to prove: 10n = 10 × 10 × · · · × 10
n
(3)
for all whole numbers n > 0.e.
n
We need to clarify this situation by proving that these two numbers are the same. of 10 in the expanded form of a number indicates precisely that the associated digit is the (k + 1)-th one from the right. which immediately signals that it stands for 30000 and not 3. Thus the expression 3 × 104 tells us that 3 is in the ﬁfth place (from the right) of 830159. There is one aspect of the expanded form that may trouble you: why use the cumbersome notation of 5 × 101 and 9 × 100 in the expanded form of 830159 instead of just 5 × 10 and 9 ? The answer is: it all depends on what we want. i. . say k. and 2 × 106 indicates the position of 2 as the seventh digit from the right of 2070040. the less rigid notation of 5 × 10 and 9 is suﬃcient for ordinary purposes. When absolute conceptual clarity is called for. The ﬁrst one is that some of you may have encountered another deﬁnition of 10n for a nonzero whole number n as 10 × 10 × · · · × 10.. so usually we just write: 830159 = (8 × 105 ) + (3 × 104 ) + (1 × 102 ) + (5 × 10) + 9 . The clear location of a digit in the expanded form of a number will turn out to be very helpful in understanding all the arithmetic algorithms in §3.

2. (3) may be restated as 10n = 10[n] for all whole numbers n > 0. why is 102 = 10[2]? This is because 102 = 10 × 101 (by ﬁrst equality of ( ) with n = 1) = 10 × 10[1] (by 101 = 10[1]) = 10[2] (by second equality of ( ) with n = 1)
. say n = 1. 0) and 10[n] are “built up” in exactly the
n
n
same way. . 000 = 213. 0023 is the same as 23.070 by tagging the 5 zeros of 100.
A second loose end we should tie up is a fact that many of you probably take for granted. 10n+1 = 10 × 10n and 10[n + 1] = 10 × 10[n]. The reason is twofold: (a) It is true for n = 1.1 Place Value
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Activity: Prove that (3) is true for n = 1. 3. but it could only be because you are used to thinking of n as a small number.
n
then both symbols 10 (= 1 00 . Next.g. . in the sense that for any nonzero whole number n. e.133.e.000 to the right of 2.070. 2..
You may think that (3) is obvious. ( )
Note that using the temporary notation. 307. 133.000 (say) results in a number that is obtained from 2. 000 The last loose end we wish to address is the issue of the implicit zeros in front of any whole number. 2. i.133. it is quite obvious that both 101 and 10[1] are equal to 10 and therefore equal. that for any number. 070 × 100. 3. and (b) if we introduce the temporary notation that 10[n] = 10 × 10 × · · · × 10.
Let us ﬁrst make use of (a) and (b) to prove this before we prove (a) and (b) themselves. That is.070. multiplying it by 100.
Let us ﬁrst explain why (3) above is true. For n = 1.133. namely.. 000. 2. But what about n = 123457321009? Can you even make sense of (3) in that case? Thinking about these questions will perhaps convince you that we should spend some time ﬁnding out why (3) is true for all values of n.

it is not at all clear why (4) should be true. If n is a big number. without (3). can you be absolutely certain that adding 100 . This has obvious implications. . 0 (n zeros) to itself 100 . e. . and m = 68739. we can dispatch (4) easily enough. 00 (68739 zeros) times would result in the number 1000 .. Then: 10m × 10n = 10 × 10 × · · · × 10 × 10 × 10 × · · · × 10
m n
= 10 × 10 × · · · × 10
m+n
= 10m+n So we see that (4) is true. Remember. . which we could do without any diﬃculty. . 10m is by deﬁnition the number 100 . so the left side of (4) means (by the deﬁnition of multiplication in (2)) adding 100 . by using (3). . such as 21334658.
4
But see the discussion of the associative law of multiplication in §2. 10m × 10n = 10m+n (4)
Because (3) is available to us. .1 Place Value
17
10 × 10 × · · · × 10 (n times).
. if m and n are any whole numbers. . . . If m = 0 or n = 0. there is nothing to prove. 00 (m zeros) times. 00 ((68739 + 21334658) zeros) ? So the key point of the proof of (4) is that. n > 0. 0 (21334658 zeros) to itself 100 . 101 × 103 102 × 104 103 × 104 104 × 105 = = = = 104 106 107 109 . . so we may assume m.g. .
This suggests that in general.4 It is a point well worth making that. 00 (m zeros). the evaluation of the product 10m × 10n is reduced to the counting of the number of copies of 10’s. .

and the age of the universe as 14 × 109 years. which is 5. then N × 10k = the whole number obtained from N by attaching k zeros to the right of the last digit of N . 536 × 103 . 000 = 213.696. 000 = { (2 × 106 ) + (1 × 105 ) + (3 × 104 ) +(3 × 103 ) + (7 × 101 ) } × 105 = (2 × 106 × 105 ) + (1 × 105 × 105 ) + (3 × 104 × 105 ) + (3 × 103 × 105 ) + (7 × 101 × 105 ) = (2 × 1011 ) + (1 × 1010 ) + (3 × 109 ) + (3 × 108 ) + (7 × 106 ) (using (4)) = 213. We can now write this number as 5. incidentally. which will be discussed presently in the next section. 2. 000 The same reasoning of course allows us to write 213.307. The above reasoning is suﬃciently general to show that if N is any nonzero whole number and k is any whole number.000 gives a number which can be obtained from 2.070 times. appearance notwithstanding.070 by adding the 5 zeros of 100.070.133. This is. the left side means adding 100.
5
.000. 133. 070 × 100. The last 5 digits of the resulting number will be 0’s. That is. but why are the beginning digits exactly 2133070 ? The reason is: 2.070 by 100.5 but it will hardly be the last.000 miles. 133. 000. 000. In a similar way.000.000 to itself 2.133.
The discerning reader would have noticed that the above derivation made implicit use of the distributive law. 000 Note that. astronomers use as their unit of measurement a light year.000 more simply as 213. this fact is again not obvious! According to (2). Furthermore.133. 696 × 106 miles. 307 × 106 . to be sure. 070 × 100. 865. 307. we can now write the number of seconds in a year as 31.133.000 to the right of 2.1 Place Value
18
We are now in a position to explain the fact that multiplying 2.865. our ﬁrst substantive application of the expanded form of a number. 307.

one ﬁnds: How many are in 4 groups of 6? You can use multiplication to solve the problem.6 Write each of the numbers 6100925730. Exercise 1. we deal with the issue of why the number 830159 is the same number as 0830159. Exercise 1. Then in the text proper.1 Imagine you have to explain to a fourth grader that 43 × 100 = 4300.1 Place Value
19
Finally. How would you do it? Exercise 1.4 What number should be added to 58 × 104 to get 63 × 104 ? What number should be taken from 52 × 105 to get 48 × 105 ? Exercise 1. 000. 10k > m×10k−1 for any single-digit number m.8 The following is the introduction to the concept of multiplication taken from a third grade textbook. Exercise 1. etc. called a product.953? Exercise 1. This is very easy to explain because by the expanded form of a number. 000830159 = 0 × 108 + 0 × 107 + 0 × 106 + 8 × 105 + 3 × 104 + 1 × 102 + 5 × 101 + 9 × 100 = 830159 .214. and 7420000659 in expanded form.2 Imagine you have to explain to a ﬁfth grader why 48 × 500.214.722 to get 986. 000.953 to get 88. Exercise 1. How would you do it? Exercise 1.919? Explain.833 or 764. Use cubes to model the problem and record the answer to the problem:
. and is the same number as 000830159. This observation is conceptually important in the understanding of the various algorithms in §3. 2300000000. 000 = 24. 000. 000.7 Show that for any nonzero whole number k.3 What number should be added to 946.019.5 Which is bigger? 4873 or 12001? 4×105 or 3×106 ? 8×1032 or 2 × 1033 ? 4289 × 107 or 1011 ? 765.722? What number should be added to 68.927. On the side of the page is the Vocabulary of the Day: multiplication an operation using at least two numbers to ﬁnd another number. product the answer in multiplication.

The key point of such a discussion is always that two collections of numbers which look superﬁcially diﬀerent are in fact equal. Write down your reaction to the appropriateness of such an introduction. you can also write a multiplication sentence. and this equality is indicated by the ubiquitous equal sign “=”.2 The Basic Laws of Operations
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Number of groups 6 Further down:
| Number in each group × 4
| Product = 24
If Helena practices singing 3 hours each day for a week. 7 × 3 = 21. why 23 + 79 = 79 + 23. so the meaning of “=” is both simple and unequivocal: two numbers a and b are said to be equal if we can verify by counting that they are the same number. or why (47 × 4) × 5 = 47×(4×5). e. how many hours will she practice altogether? Find: 3 × 7 There is more than one way! Method 1: You can use repeated edition to solve the problem. there is no reason for us to learn everything about this symbol all at once! We are starting from ground zero.. The most important thing to remember is that while the meaning of the equal sign does get more sophisticated as the mathematics gets more advanced. 3 + 3 + 3 + 3 + 3 + 3 + 3 = 21 Method 2: When the groups are equal. Because the equal sign is one of the sources of confusion in elementary school.g. let us ﬁrst deal with this symbol. and compare your view with those of others’ in your class.
2 The Basic Laws of Operations
This section discusses the basic laws which govern the arithmetic operations on whole numbers.
. the whole numbers.

some teachers probably explain this shortcut to their second graders by saying you just “ﬂy over” the middle number and add the two at both ends. The equality of two fractions or two decimals will have to be more carefully explained (see §1 of Chapter 2 and §§1 and 3 of Chapter 4). 070 × 100. the simple problem of addition: 12 + 25 + 18. Yet we are going to spend the next twenty pages discussing exactly these laws without any apology. Because these laws inﬁltrate every aspect of arithemtic operations. Take. A quick way to compute this sum is to add 12 and 18 to get 30. When we deal with fractions (Chapter 2) and decimals (Chapter 4). There are at least three reasons. 133. so this is what 4 + 5 = 2 + 7 means. It merely means: check the numbers on both sides of the equal sign by counting to verify that both sides yield the same number. So be sure to explain to your students — again and again if necessary — that the equal sign between two collection of whole numbers does not signify “do an operation to get an answer”. whereas we count 2 objects and then 7 more and also get 9. 000. let me just cite two of many instances where we have used them in §1 in an “underhanded” manner: the proof that 2. and the distributive law. then obviously we can no longer just count. 4 + 5 = 2 + 7 because we count 4 objects and then 5 more and get 9. Now. 307.2 The Basic Laws of Operations
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For example. for instance. as we shall explain below (see the discussion above (10)). and then add 30 to 25 to get 55. presumably they e believe you deserve better.
. sometimes implicitly without your being aware of them. your awareness of their presence would help you avoid making incorrect pronouncements. but this would be incorrect because by convention. The ﬁrst is that they are used everywhere. and the proof of (4) (that 10m × 10n = 10m+n ) made implicit use of the associative law of multiplication. 000 = 213. You are entitled to know why. 000 implicitly made use of the distributive law. To bring home this point. Now we come to the main concerns of this section: the associative laws and commutative laws of addition and multiplication. These are without doubt among the most hackneyed items you have ever come across in mathematics. Your textbooks mention them with a sense of noblesse oblig´ and can’t wait to get it over with.

these computations can all be done eﬀortlessly in one’s head so that the answers are.. or 25 + 18. To add or multiply two number by “ﬂying over” another number is not a permissible move in mathematics. As a teacher. For this reason. it is understood that one only adds or multiplies two neighboring numbers at a time: e. §7. 871690. is that these laws play a central role in this monograph. we take the opportunity to clarify these foundational matters once and for all. Yet with a judicious application of the basic laws. if L. and §3 of Chapter 5. but are especially prominent in §§3. and the commutative and associative laws of multiplication. the deﬁnition in (2) of §1 implies that (b) must be computed by adding 6572 to itself 10. They weave in and out of the ﬁve chapters. 12 + 25.g. (L + M ) + N = L + (M + N ) M +N = N +M (5) (6)
These are nothing more than summaries of our collective experiences with
.000.2 of Chapter 2. respectively. respectively. A ﬁnal reason. the two basic laws are the associative law and the commutative law. or 4 × 17. and the most substantive one for our purpose. For example. For addition. you have to be ready with the correct mathematical explanations for such phenomena when the occasion calls for it.4 of Chapter 1.3–3.000. 000 times. and 5300. because the apparent “ﬂying over” can be precisely justiﬁed by the commutative and associative laws of addition. or 17 × 25. See the later discussions in this section for the explanations. 000.720. 65. A second reason is that knowing these basic laws can be helpful. These state that.2 The Basic Laws of Operations
22
in any expression involving arithemtic operations such as 12 + 25 + 18 or 4 × 17 × 25. M and N are whole numbers. Consider the following simple problems: (a) (87169 × 5) × 2 = ? (b) 107 × 6572 = ? (c) 4 × ([25 × 18] + [7 × 125]) = ? These computations can be tedious if taken literally. But the possibility of computing 12 + 25 + 18 as (12+18)+25 = 30+25 = 55 or 4×17×25 as 17×(4×25) = 17×100 = 1700 will turn out to be entirely correct.

M = 4 and N = 7:
2 4 7
2
4
7
It is obvious from the pictures that whether we combine 2 and 4 ﬁrst and then combine the sum with 7. As to the commutative law (6). and we take their truth as an article of faith. What (5) says is that these two numbers are the same number. We pause to comment on the equal sign in the context of the associative law (5) to reinforce the deﬁnition of this symbol.
.6 For illustrations of some of these experiences. then continue the counting to include the group consisting of M objects and N objects. What (5) asserts is that the two numbers (L + M ) + N and L + (M + N ) are the same. the pictorial evidence for the truth of (5) can be similarly obtained for other numbers. and then continue the counting to include the next N objects. we begin with L objects. For the right side. either way we get 8 objects:
3 5
5
3
Again. thereby getting a second number. we get one number. we count the left side by ﬁrst counting the group consisting of L objects and M objects.
6
In the proper axiomatic setup. think of the addition of two numbers as combining two groups of discrete objects. Needless to say.2 The Basic Laws of Operations
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whole numbers. both of (5) and (6) can be proved as theorems. the pictorial evidence is not dependent on the particular choice of M = 3 and N = 5 and would hold equally well for other numbers. Then the following pictures describe the validity of the associative law (5) for L = 2. or combine 2 with the sum of 4 and 7. In other words. the following pictures show that whether we combine 3 objects with 5 objects or 5 objects with 3 objects. we get 13.

and ﬁnally add ((l + m) + n) to p. then add the result to n. however. as desired. and N = p and reading (5) from right to left. For example. because it could mean either (4 + 3) + 7 or 4 + (3 + 7) . M = n. Consequently. Once this is noted. we get ((l + m) + n) + p = l + ((m + n) + p). (5) leads to the conclusion that we don’t need to use parentheses in writing the sum of any three whole numbers l + m + n . M = m.. we obtain m + (n + p) = (m + n) + p.g. Letting L = (l + m). ()
Now let L = l.) But (5) tells us that there is in fact no ambiguity because the two ways of adding are the same. we obtain ((l + m) + n) + p = (l + m) + (n + p). In general then. ()
. the meaning of l + m + n + p is also unambiguous without the use of parentheses. Putting ( ) and ( ) together. it should not be surprising that we can draw the same conclusion about the sum of any four whole numbers l + m + n + p. Let us explain the preceding equality of four numbers by using (5) repeatedly. letting L = m.2 The Basic Laws of Operations
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One consequence of the associative law of addition is that it clears up the meaning of such common expressions as 4 + 3 + 7. Substituting this value of m + (n + p) into the right side of the preceding equation. we obtain: (l + m) + (n + p)) = l + ((m + n) + p). This triple addition is a priori ambiguous. and N = p in (5). let us show: ((l + m) + n) + p = l + ((m + n) + p). where the convention regarding parentheses is to do the innermost parentheses ﬁrst and then systematically work one’s way out. only neighboring numbers are added. Finally. 4 + 3 + 7 . then the right side of ( ) becomes (l + m) + (n + p) = l + (m + (n + p)). by convention. and N = (n + p) in (5) again. Thus ((l + m) + n) + p means: add l to m. (Note again that adding 4 to 7 before adding the result to 3 is not an option because. e. M = n.

an argument for the general case (which should be more diﬃcult) turns out to be more understandable than the special case. m = 2. y. if we let l = 5. For example. regardless of the order of the addition of the numbers. by resorting to formal abstract reasoning.
.2 The Basic Laws of Operations
25
Because this kind of abstract. say 26 of them {a. we have already made implicit use of this fact. we can unambiguously write a + b + c + ··· + y + z without the use of any parentheses and. m and n. there are many other examples of using the associative law of addition without mentioning it. z}. i. The same can be said of the later expression (in §1) that 31536000 = 30000000 + 1000000 + 500000 + 30000 + 6000.7 the result will be the same. formal reasoning looks so facile and believable. · · · . You cannot fail to notice at this point that in the very ﬁrst sum of §1. it is far from obvious why they turn out to be equal at the end: ((5 + 2) + 8) + 1 = (7 + 8) + 1 = 15 + 1 = 16 5 + ((2 + 8) + 1) = 5 + (10 + 11) = 5 + 11 = 16 This is the ﬁrst instance that we come across the paradoxical situation that. n = 8 and p = 1 and compute both ((5 + 2) + 8) + 1 and 5 + ((2 + 8) + 1) directly to compare the results step-by-step.e. The same reasoning shows that given any collection of numbers. Let me therefore make sure you are aware that there is substance beneath the smooth surface of formalism. I want to assure you that this will not be the last instance where this happens. Needless to say.. For example. there is the danger that you would take it for granted. 20000 + 500 + 40 + 1. if we have three numbers l. b. c. then all six expressions l+m+n m+n+l
7
l+n+m n+l+m
m+l+n n+m+l
Don’t forget the convention that one can only add two neighboring numbers at a time. Essentially the same comments apply to the commutative law (6).

If we have ﬁve numbers 2. 5 11. ((4 + 5) + 11) + (2 + 3) = (9 + 1) + 5 = 20 + 5 = 25. that l + m + n = n + m + l by applying (6) repeatedly to two numbers at a time: l+m+n = m+l+n = m+n+l = n+m+l Of course the same is true of the addition of any collection of whole numbers: the order of appearance of the whole numbers in any ﬁnite sum is unimportant. We should emphasize that argument of this sort is primarily for your beneﬁt as a teacher and should not be taken as a statement that good teaching means always explaining details of this kind. In the fourth or ﬁfth grade. Observe ﬁrst of all that the associative law of addition is already used in writing 12+25+18 without the use of parentheses. 3.2 The Basic Laws of Operations
26
are the same. we are arguing that 12 + 25 + 18 = (12 + 18) + 25. and there is no need to do any of the other 119 sums because they would all be equal to 25. you undoubtedly appreciate one aspect of the laws (5) and (6): they are tremendous labor saving devices. But we only have to do one of them. This then justiﬁes the writing of 2 + 5 + 4 + 11 + 3 = 25. By now. for instance. there will be occasions for you to introduce this kind of
. we give the proof: 12 + 25 + 18 = 12 + (25 + 18) = 12 + (18 + 25) = (12 + 18) + 25 (commutative law) (associative law)
Exactly as claimed. it may be time to tie up the loose end left open earlier as to why it is permissible to ﬁrst add 12 + 18 in the sum 12 + 25 + 18. (Notice that we have already made use of the fact that the use of parentheses is not needed for the addition of three numbers !) Let us show. for instance. you should be ready with an explanation. 4. Before proceeding further. we may insert parentheses any which way we wish. as a teacher. So with this understood. In detail. then there are 120 ways of adding them (try it!). say. What it does say is that.

We have two similar laws.000 to itself 6572 times. First. by the associative law (7). Next. 107 × 6572 is adding 6572 to itself 10. The equality of these two products may not seem important to you because you have taken it for granted. but the right side is 65. The way we have just dealt with (a) and (b) aﬀords an excellent opportunity to remind you of the importance of using deﬁnitions exactly as given. But if you do not begin to take note (7) (8)
.720. This disposes of (a).000 by the conclusion we arrived at near the end of §1.000. Before discussing the empirical evidence behind (7) and (8).000. (LM )N = L(M N ) MN = NM where we have made use of the notational convention: when letters are used to stand for numbers. at least. But a teacher must exercise good judgment in not overdoing anything. Next we turn to the multiplication of whole numbers as deﬁned in (2) of §1. for you to equate in your minds without thinking that 107 × 6572 = 6572 × 107 and then conclude that the result is 65. M and N . They look at ﬁrst glance.2 The Basic Laws of Operations
27
reasoning to your students. Since 5×2 is 10.720. This then disposes of (b). It would have been very easy. the multiplication sign is omitted so that “M N ” stands for “M × N ”. What we are doing here is to intentionally bring the underlying reason (the commutative law (8)) for the validity of 107 × 6572 = 6572 × 107 to your attention: by deﬁnition.000 times. the triple product is immediately seen to be equal to 871690. to the eﬀect that N × 10k = the whole number obtained from N by attaching k zeros to the right of the last digit of N . the associative law of multiplication and the commutative law of multiplication: for for any whole numbers L. to be quite diﬀerent animals.000.000.000. 107 ×6572 = 6572×107 by the commutative law (8). gently. (87169×5)×2 is equal to 87169×(5×2). we can demonstrate their power by applying them to problems (a) and (b) posed at the beginning of this section. whereas 6572 × 107 is adding 10. for example.

by using 3-dimensional rectangular arrays of dots to represent triple products of whole numbers. you would not be able to follow the reasoning in later discussions when the direction of the discussion is determined by (8).2 The Basic Laws of Operations
28
of the role of (8) in arithmetic.g.4 when we show that the two interpretations of division yield the same result. Here is the kind of pictorial evidence that is easily available: Activity: Verify by direct counting that (8) is valid for M and N between 1 and 5 inclusive (in symbols: 1 ≤ M. in §3. For example. the product 5 × 3 would be represented by the area of a rectangle with “vertical” length 5 and “horizontal” length 3:
. N ≤ 5). by using rectangular arrays of dots to represent multiplication of whole numbers as in the discussion below (2) in §1. the discussions of fractions and decimals in Chapters 2 and 4). N between 1 and 5 inclusive (in symbols: 1 ≤ L. M. Activity: Verify by direct counting that (7) is valid for L. 3 × 5 is represented as a rectangle with “vertical” length 3 (corresponding to the ﬁrst number) and “horizontal” length 5 (corresponding to the second number):
(9)
According to this description of the area model.. As with the corresponding laws for addition. the representation of multiplication by dots would be inadequate and it would be more suitable to use a area model. For later needs in this monograph (cf. M . N ≤ 5). one may regard (7) and (8) as summaries of empirical experiences and accept them on faith. e.

like other manipulatives. they have the same area. Most likely you have used this manipulative in your classroom to facilitate the learning of multiplication. so that the area of the rectangle in (9) is 15 because precisely 15 unit squares tile (or pave) the rectangle. this discussion is independent of the choice of the numbers 3 and 5 and would be the same for any two numbers m and n. by convention. As usual. We would like to add a passing comment that while Base Ten Blocks.
. students should not be allowed to become dependent on it. It is worthwhile to point out that the area model of multiplication provides the mathematical underpinning of the manipulative Base Ten Blocks. not manipulate manipulatives. nor will it be needed. we will present yet another interpretation of the multiplication of whole numbers that will be important when we come to fractions.
In §4. can be helpful. we agree to let the area of the unit square (the square with each side equal to 1) to be just 1. We cannot give this interpretation here because it involves a deeper understanding of what a “number” is.
A word about “area” would be appropriate: A detailed discussion of this concept will not be attempted here.2 The Basic Laws of Operations
29
Because these two rectangles can be obtained from each other by a 90◦ rotation. except to mention that. This then give a pictorial “explanation” of 3 × 5 = 5 × 3. The real challenge in a mathematics classroom is still to learn the mathematics.

For example. and we know that
. m and n can be represented as the volume of a rectangular solid. Of course.2 The Basic Laws of Operations
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The product of three whole numbers l. as in the case of addition. We remark as in the case of area that we shall not go into the precise deﬁnition of volume but will only use it in an intuitive way. For example. and 2 of the preceding argument can be replaced by any triple of numbers in (7). so that 3 × (5 × 2) is the volume of the solid of height 3 built on the rectangle with “verticle” length 5 and “horizontal” length 2:
¨ ¨ ¨ ¨¨ ¨¨ ¨¨ ¨ ¨¨ ¨¨ ¨¨ ¨ ¨ ¨ ¨¨ ¨ ¨¨ ¨ ¨ ¨¨ ¨¨ ¨ ¨ ¨¨ ¨¨
The equality of the volumes of these two solids — because one is obtained from the other by a rotation in space — is then the pictorial evidence for the truth of (3 × 5) × 2 = 3 × (5 × 2). Again. (3 × 5) × 2 is the volume of the rectangular solid of height 2 built on the rectangle in (9) above:
¨ ¨ ¨ ¨ ¨ ¨ ¨¨ ¨¨ ¨¨ ¨¨ ¨¨ ¨¨ ¨ ¨¨ ¨¨ ¨¨ ¨¨ ¨¨ ¨ ¨¨ ¨ ¨ ¨ ¨ ¨ ¨ ¨¨ ¨¨ ¨ ¨ ¨¨ ¨ ¨
Similarly. 5. the concrete numbers 3. the previous discussion of the associative law and commutative law for addition ((5) and (6)) applies to multiplication verbatim. there are more general versions of (7) and (8) for arbitrary collections of numbers. 3 × (5 × 2) = (5 × 2) × 3 by the commmutative law (8).

mlpn = plnm: (7 × 3) × (2 × 4) = 21 × (2 × 4) = 21 × 8 = 168 (4 × (7 × 2)) × 3 = (4 × 14) × 3 = 56 × 3 = 168 . implicitly made use of the associative law of multiplication. namely. say l. we have 10m = 10 × 10 × · · · × 10. we also know that a product of m + n numbers can be written without the use of parentheses. Properly speaking. let us use four explicit numbers — say l = 7. m = 3. 10n = 10 × 10 × · · · × 10. This shows 10m × 10n = 10m+n . Similarly. To drive home the point that ﬁrst surfaced in connection with the associative law of addition. for example.
. n = 2 and p = 4 — to illustrate the nontrivial nature of. n.2 The Basic Laws of Operations
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the product of any collection of numbers can be written unambiguously without the use of parentheses and without regard to order. 10m × 10n = 10m+n . Therefore. m. note that none of the intermediate steps of the two computations look remotely similar. its proof should go as follows: Because we may write a product of m numbers without the use of parentheses. p. and yet miraculously the ﬁnal results are identical. their product can be written in any of the following 24 ways: lmnp mlnp nlmp plmn lmpn mlpn nlpm plnm lnmp mnlp nmlp pmln lnpm mnpl nmpl pmnl lpmn mpln nplm pnlm lpnm mpnl npml pnml
and all of them are equal to ((lm)n)p. Consequently. Thus. 10m × 10n = 10 × 10 × · · · × 10. At the risk of harping on the obvious. Now we are in a position to point out in what way the proof of (4) in §1. for any four numbers.
m+n
and the right side is just 10m+n .
m n
10m × 10n = (10 × 10 × · · · × 10) × (10 × 10 × · · · × 10)
m n
However.

the distributive law (10) is nothing more than a summary of common sense. Thus 3(2 + 4) = (3 × 2) + (3 × 4). Pictorially. c and d. we have: Activity: Directly verify (10) using rectangular arrays of dots to represent multiplication for 1 ≤ L. and L. the distributive law generalizes to more than three numbers. N = 2 and L = 4. It states that for any whole numbers M . Clearly. For example: m(a + b + c + d) = ma + mb + mc + md for any whole numbers m. N . 3 × 2 is the area of the left rectangle and 3 × 4 is the area of the right rectangle. we multiply the numbers M N and M L ﬁrst before adding. Again. 3. then 3(2 + 4) is the area of the following big rectangle: (10)
On the other hand. One can also use the area model: if M = 3. as follows: m(a + b + c + d) = = = = m({a + b} + {c + d}) m{a + b} + m{c + d} {ma + mb} + {mc + md} ma + mb + mc + md
. the distributive law connects addition with multiplication. a. M. and 4 are replaced by other triples of numbers.2 The Basic Laws of Operations
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Finally. N ≤ 5. the essence of this picture is unchanged when 2. M (N + L) = M N + M L Recall in this connection the convention that in the expression M N + M L. b. like the other laws we have discussed so far. Again. This can be seen by applying the distributive law (10) twice.

e. and 4 × 1325. but when (35 × 72) + (35 × 29) is given instead. the latter skill may be the more critical of the two. 7 × 125. (35 × 72) + (29 × 35) involves two multiplications ( 35 × 72 and 29 × 35 ) and one addition. Therefore: 4 × ( 25 × 18 + 7 × 125) = = = = ) 4 × (25 × 18 + 25 × [7 × 5]) 4 × 25 × (18 + [7 × 5]) (distributive law) 100 × 53 = 53 × 100 5300. Despite its obvious importance. Therefore. they fail to see that it is equal to 35 × (72 + 29). to recognize (35 × 72) + (29 × 35) = 35 × (72 + 29) is to be able to achieve a simpliﬁcation. M N + M L = M (N + L)). it seems to be the least understood among the three laws and for many. the distributive law is the glue that binds addition + and multiplication × together. given 35 × (72 + 29). it is preferable to multiply as little as possible and therefore preferable to compute 35 × (72 + 29) rather than (35 × 72) + (29 × 35). we would be looking at calculating three multiplications if we don’t appeal to the distributive law: 25 × 18.. A common mistake in connection with the distributive law is to remember (10) only in the form of “the left side of (10) is equal to the right side” (i. In practice.
It should be clear that the whole computation can be done by mental math. In other words. Because multiplication is in general more complicated.. To further pin down the last idea. We urge you to spare no eﬀort in learning to use it eﬀectively. and there is an mathematical reason for this. most people recognize that it is equal to (35 × 72) + (35 × 29). so that 7 × 125 = 7 × 5 × 25 = 25 × [7 × 5]. On the other hand. whereas 35 × (72 + 29) involves only one multiplication and one addition.
.e. As mentioned earlier. we bring closure to this section by doing problem (c) posed near the beginning: 4 × (25 × 18 + 7 × 125) = ? We know 125 = 5 × 25. where 1325 = 25 × 18 + 7 × 125. M (N + L) = M N + M L) without realizing that (10) also says “the right side of (10) is equal to the left side” (i.2 The Basic Laws of Operations
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Observe that each step in the preceding calculation depends on the earlier discussion of the legitimacy of writing a + b + c + d without parentheses. a ﬁrm grasp of this law proves elusive. In terms of the preceding example. Spending more time with some of the exercises at the end of this section may be the answer.

if there is a nonzero whole number c so that a+c = b. so 7 < 12. We began the discussion of order among whole numbers (i. we have to go 5 more steps before we get to 12. Given a < b. so 7 + 5 = 12. we wish to express this deﬁnition diﬀerently: The statement a < b is the same as the statement: there is a nonzero whole number c so that a + c = b. Conversely. 2. Before giving the reason for this assertion. At this juncture. 3. we need to go (let us say) c more steps before getting to b. etc. after we get to a. if we know 7 + 5 = 12. We can now conclude that discussion. it is time to introduce the number line in order to give a geometric interpretation of inequalities. Fix a straight line and designate a point on it as 0. so that in the counting of the whole numbers.. Thus the whole numbers are now identiﬁed with these equally spaced points on the right side of the line. We explicitly call attention to the fact
. we know a < b. with c > 0. we have to count c more objects before we get b objects. Recall that given two whole numbers a and b. then we must go 5 more steps from 7 before we get to 12. . 7 < 12 means that in counting from 7. we know by deﬁnition that a precedes b. again as on a ruler. This implies a + c = b. It is convenient to single these points out by markers (notches). suppose a + c = b is given. For example. the inequality a < b is deﬁned to mean that in the counting of the whole numbers starting with 0.. and to identify the points with the markers. Conversely.e. 2. then there is a nonzero whole number c so that a + c = b. In concrete situations. This is a piece of mathematical terminology that signiﬁes that both of the following statements are true: If a < b. .2 The Basic Laws of Operations
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Moral: Be sure you know M N + M L = M (N + L). we explain what is meant by the two conditions being the same. then after counting a objects. both statments are quite obvious. as on a ruler. The general reasoning is not much diﬀerent. the number a precedes b. then a < b. which whole number is bigger) in §1. For the convenience in logical arguments. Conversely. So by the deﬁnition of “smaller than”. . and c is not zero. . mark oﬀ equally spaced points and call them 1. 4. To the right of 0. 1.

(Until Chapter 5. as shown below. Do it both numerically as well as on the number line. or a = 86. Activity: Verify the last statement about c for some concrete numbers such as a = 8. This follows immediately from the way the whole numbers are positioned on the number line. The following facts about inequalities are well-known: a + b < a + c is the same as b < c a = 0. Here is a pictorial representation of the situation: a<b a b
A further geometric interpretation can be given: given a < b as shown. Activity: Convince yourself that 2a < 2b is the same as a < b.
.
It follows from the way the number line is drawn that two whole number a and b satisfy a < b precisely when the position of a on the number line is to the left of that of b.2 The Basic Laws of Operations
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that the counting of the whole numbers (as was done in §1) corresponds to the progression of the markers to the right of the line starting with the initial marker 0. then c is precisely the number of markers between a and b. suppose a + c = b. b = 21. b = 95. and that 3a < 3b is the same as a < b. ab < ac is the same as b < c a < b and c < d implies a+c<b+d a < b and c < d implies ac < bd
(11)
We will only give the explanation for the second assertion (the most diﬃcult of the four) and leave the rest as exercises. we will have no need for the part of the line to the left of 0): 0 1 2 3 4 5 etc.

Given a = 0. so that ab + al = ac again by the distributive law.
. (ii) (54 + 69978) + 46. so al > 0 and therefore ac < ab. Exercise 2.. and there will be a nonzero whole number l so that c + l = b. (i) Let A1 stand for “put socks on” and A2 for “put shoes on”. It follows that (C) is the only possible conclusion. We have therefore eliminated (A). then c < b. which implies ac + al = ab. Show that no matter what the number n may be. b < c implies ab < ac. then ab = ac. First we prove that ab < ac implies b < c. and then A1 ”.2 The Basic Laws of Operations
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Now the proof of the second assertion. and similarly let A2 ◦ A1 be “do A2 ﬁrst. Exercise 2. (B) c < b. M ) (viii) (([125 × 24] × 674) + ([24 × 125] × 326) .1 Elaine has 11 jars in each of which she put 16 ping pong balls. i. Next.2 Before you get too comfortable with the idea that everything in this world has to be commutative. consider the following. let B1 (k) be the number obtained by adding 2 to k. all we need to do is to show that both (A) and (B) are impossible. the last equation means ab < ac.) for each step: (i) 833 + (5167 + 8499). and (vi) 64 × 125. Hence (B) is also eliminated. (v) (4 × 4 × 4 × 4) × (5 × 5 × 5 × 5 × 5). Convince yourself that A1 ◦ A2 does not have the same eﬀect as A2 ◦ A1 . Thus a(b + l) = ac. If (B) holds. Because al is nonzero. We know ahead of time that only one of the three possibilities holds. we prove the converse. (ii) For any whole number k. and let A1 ◦ A2 be “do A1 ﬁrst. Now b < c implies that b + l = c for some nonzero l. (iv) (58679 × 762) + (58679 × 238). One day she decided to redistribute all her ping pong balls equally among 16 jars instead. Exercise 2. (iii) (25 × 7687) × 80. and then A2 ”. (vii) (69 × 127) + (873 × 69). But both a and l are nonzero. which is contrary to the assumption that ab < ac. There are three possibilities between the whole numbers b and c : (A) b = c. This is also contrary to the assumption that ab < ac. This completes the proof. by the distributive law.3 Find shortcuts to do each of the following computations and give reasons (associative law of addition? commutative law of multiplication? etc. as desired. B1 (B2 (n)) = B2 (B1 (n)). and B2 (k) be the number obtained by multiplying k by 5. It follows that a(c + l) = ab. How many balls are in each jar? Explain. Now if (A) holds. and (C) b < c. In order to show that (C) is the correct conclusion.e.

where k and are nonzero whole numbers. 2 × 6 × 4 = (2 × 6) × 4
. contrary to what some people would have you believe. where as usual x2 means xx and y 2 means yy.4 Prove the remaining three assertions in (11). Exercise 2. abstract gestures. Explain why (x − y)(x + y) = (x − y)x + (x − y)y. You can use the associative property to multiply three factors. (iii) Explain why (x + y)(x + y) = x2 + 2xy + y 2 . you will be allowed to use only 8 digits!). What this exercise tries to do is.10 Let x and y be two whole numbers. How many digits can the number mn be? List all the possibilities and explain. They have practical applications too. but you have to calculate 856164298 × 65.
Exercise 2.11 The following is how a fourth grade textbook introduces the associative law of multiplication. The grouping of the numbers does not affect the answer. rather.2 The Basic Laws of Operations
37
The purpose of the last exercise is not to get you obsessed with tricks in computations everywhere. Exercise 2.7 Let m and n be a k digit number and an digit number. Explain. Discuss an eﬃcient method to make use of the calculator to help with the comutation. but they are not the main goal of a mathematics education. Exercise 2.8 Suppose you have a calculator which displays only 8 digits (and if you have a fancy calculator. He buys six different styles from each company and gets each style in 4 different colors. Step 1: Use parentheses to show grouping.6 Let m and n be a 3 digit number and a 2 digit number. Exercise 2.9 Let x and y be two whole numbers. Tricks are nice to have.5 Prove that both assertions in (11) remain true if the strict inequality symbol “<” is replced by the weak inequality symbol “≤”. (i) Explain why (x + y)(x + y) = x(x + y) + y(x + y). Do the same for 376241048 × 872. How many yo-yos does he buy in all? Find 2×6×4 to solve. to make you realize that the basic laws of operation discussed in this section are more than empty. Exercise 2. Ramon buys yo-yos from two companies. respectively. Exercise 2. Can mn be a 4 digit number? 5 digit number? 6 digit number? 7 digit number? Exercise 2. (ii) Explain why (x + y)(x + y) = xx + xy + yx + yy. respectively.

is a marvel of human invention. especially the multiplication algorithm and the long division algorithm. but have been incorporated into the algorithms by various countries and even diﬀerent ethnic groups.
. we should make clear that there is no such thing as the unique standard algorithm for any of the four operations +. (2 × 6) × 4 = (6 × 2) × 4 = 6 × (2 × 4) = 6 × 8 = 48 Write down your reaction to such an introduction. At the outset. For this reason. the nomenclature of “standard algorithms” is eminently justiﬁed. they are. we shall concentrate on the mathematical ideas behind them as you are likely to be less familiar with these ideas. Step 3: Use the Commutative Property to change the order. we mean an explicitly deﬁned. if necessary. ×. because minor variations in each step of these algorithms not only are possible. This is not to say that the algorithms themselves — the mechanical procedures — are of no interest. One of the goals of this section is to make sure that you come away with a renewed respect for them. The underlying mathematical ideas are however always the same. −. ÷. and compare with those of others’ in your class. because computational techniques are an integral part of mathematics.3 The Standard Algorithms
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Step 2: Look for a known fact to multiply. On the contrary. 2 × 4 is a known fact. and it is these ideas that are the focus of our attention here. the conciseness of these algorithms. Furthermore.
3 The Standard Algorithms
By an algorithm. step-by-step computational procedure which has only a ﬁnite number of steps. Nevertheless. The purpose of this section is to describe as well as provide a complete explanation of the socalled standard algorithms for the four arithmetic operations among whole numbers.

The present crisis in the learning of algebra in schools would have been largely eliminated if students were properly taught the logical reasoning behind the algorithms in the upper elementary and middle schools. This is where algorithms come in: they provide a shortcut in lieu of direct counting. in mathematics. Much more is true. But what about 34. Therefore at the outset. learning does not take place without a solid foundation. Take a simple example: what is 17 × 12 ? By deﬁnition. 728 ? Because brutal counting is less than attractive for numbers of this magnitude. the eﬃciency of an algorithm — how to get the answer as simply and quickly as possible — is an overriding concern. But one could push this argument further. including abstract reasoning and symbolic manipulative skills. These are skills absolutely essential for the understanding of fractions and decimals in the subsequent chapters. a shortcut is clearly called for. Why worry about eﬃciency if pushing buttons on a calculator may be the most eﬃcient way to make a computation such as 34. There is a kind of leitmotif all through the algorithms. 609 × 549. this is 12 added to itself 17 times and one school of thought would have you count 17 piles of birdseed with 12 in each pile. 609 × 549. break the computation into n steps so that each step involves only one digit of the given number. 9 This conclusion is based on the fact that.8 A more important reason is that learning the reasoning behind these eﬃcient algorithms is a very compelling way to acquire many of the fundamental skills in mathematics. The precise meaning of this statement will be made clear with each algorithm. without a ﬁrm grasp of the place value of our numeral system and the mathematical underpinning of the algorithms. 728 ? There are at least two reasons why a strict reliance on the calculator is inadequate.3 The Standard Algorithms
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A fundamental question about these arithmetic algorithm is why you should bother to learn them.9 Please keep all this in mind when you learn the mathematics surrounding the algorithms and especially when you teach them to your students. First.
8
. it would be impossible to detect mistakes caused by pushing the wrong buttons on the calculators. but the overall idea is that those simpler single-digit computations can all be
I trust that it would be unnecessary to recount the many horror stories related to ﬁnger-on-the-wrong-button. which can be roughly described as follows: To perform a computation with an n digit number.

It is the very routineness that accounts for the fact that these algorithms get used. What one must fear is limiting one’s mastery of such procedures to only the mechanical aspect and ignoring the mathematical understanding of why the procedures are correct. More to the point. As to the non-thinking aspect of these algorithms. there is at present a perception that if anything can be done without thinking. There is however a discus-
. we should emphasize both their routine nature as well as the logical reasoning that lies behind the procedures. A teacher’s charge in the classroom is to promote both facility with procedures and the ability to reason. The ease of executing these mechanical procedures then frees up mental energy to make possible the conquest of new topics through imagination and mathematical reasoning. much of these new topics will (eventually) be once again reduced to routine or nearly routine procedures. i. The preceding discussion is about the kind of mathematical understanding teachers of mathematics must have in approaching the basic arithmetic algorithms. having such an ability in the most common mathematical situations is not only a virtue but an absolute necessity. The fact that a crucial part of mathematics rests on the ability to break down whole concepts into discrete sub-concepts and sub-skills must be accepted if one hopes to learn mathematics.3 The Standard Algorithms
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carried out routinely without thinking. There is nothing to fear about the ability to execute a correct mathematical procedure with ease. This is the very nature of mathematics. and the process then repeats itself. If we teach these algorithms without emphasizing their routine character. The last sentence calls for some comments as it runs counter to some education theories which a segment of the education community holds dear. it guarantees an easy way to get results. In turn. and no amount of philosophical discussion would change that. we would be falsifying mathematics.e. This is wrong. If mathematicians are forced to do mathematics by having to think every step of the way. then little mathematics of value would ever get done and all research mathematics departments would have to close shop.. The pedagogical issue of how to introduce these algorithms to children in the early grades is something that lies outside the scope of this monograph and needs to be treated separately. A second point concerns the uneasiness with which some educators eyes the routine and nonthinking nature of an algorithm. What is closer to the truth is that deep understanding of a topic tends to reduce many of its sophisticated processes to simple mechanical procedures. without thinking. then it does not belong in a mathematics classroom. In the teaching of these algorithms.

org/american educator/fall99/wu. we count 7 more times until we reach 11.1
An addition algorithm
Given any two numbers. y] is n for a whole number n if. — where it is understood from the notation [x. 14. n] from 0 to the whole number n to be n. say 4 and 7. here is a graphic representation of 9 + 5 = 14: 9+5 0 9 5
E
9 5
14
E
For many purposes. For any line segment [x. “Basic skills versus conceptual understanding”. y] to the left along the number line until x rests at 0.pdf
3. by sliding [x. it is more convenient to take a slightly diﬀerent point of view.aft. as follows. in the Fall 1999 issue of the American Educator. we deﬁne the length of the segment [0. y] from a point x to another point y on the number line. it is also accessible at
http://www. to ﬁnd the sum 4 + 7 is a simple matter in principle: start with 4. and that would be the sum. 7]. 7]. Simlarly. the right
. There is a graphic representation of this in terms of the number line: 4+7 0 4 4
E
11
E
7
Similarly. p. We can now deﬁne the length of more general line segments by treating the number line as an “inﬁnite ruler”.3 The Standard Algorithms
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sion of this issue concerning the addition and multiplication algorithms in the article. Denote the line segment from 0 to 7 by [0. It is natural to call 7 the length of [0. y] that y is to the right of x — we say the length of [x.

One way is to dump the content of both sacks to the ground and start counting. It can be rather trying to count 263 times starting from 4502 before getting an answer. 5] to get 9 + 5 = 14:
'
14
E
[0. Let us use a simpler example of addition to explain what is meant by the phrase in quotes.4]
[0. we count 7 more times until we reach 11”.7]
Notice that concatenating the left endpoint of [0. we can describe another way to ﬁnd the sum of 4 + 7: concatenate the two segments [0. which is what we called “the desperate act”. In principle then. 4] corresponds exactly to “start with 4. With the notion of length at our disposal. (Try it!) However. But suppose upon opening the sacks. renders such a desperate act completely unnecessary.5]
This discussion continues to be meaningful when 4 + 7 and 9 + 5 are replaced by a + b for any a + b where a and b are whole numbers. a special feature of the Hindu-Arabic numeral system. we can concatenate [0. 7] to the right endpoint of [0. one containing 34 and the other 25. Similarly. Suppose we have two sacks of potatoes. 9] and [0. Now look at 4502 + 263. namely. and we want to know how many potatoes there are altogether. 4] and [0. addition is simple. Thus:
'
11
E
[0. then the length of the concatenated segment is the sum of 4 + 7. we ﬁnd that in each sack.3 The Standard Algorithms
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endpoint y rests at n. 7] in the sense of placing them on a straight line end-to-end.9]
[0. the fact that its numerals “already come pre-packaged”. the potatoes come in bags
.

and we get 34 + 25 = 59. The following discussion of addition will be restricted to this particular version of the algorithm. Thus the total is 5 bags of 10 each. we know that there are 5 tens in the total. when this algorithm is taught in the classroom. The standard addition algorithm is nothing more than a formal elaboration of this simple idea. and adding the digits column-wise. and 2 + 3. 5 + 2 = 7. This is exactly the idea behind the addition algorithm because the number 3 in 34 — being in the tens place — signals that there are 3 tens. so there are 5 bags of 10 each). with their ones digits in the extreme right column. Observe how the addition algorithm illustrates the leitmotif mentioned near the beginning of the section: the addition 4205 + 263 is reduced to the calculation of four single-digit additions: 4 + 0. 0+6 = 6. while the sack of 25 is put in 2 bags of 10 each plus 5 stray ones. and then count the stray ones separately (4 + 5 is 9. the algorithm calls for putting a 0 there and the sum is 4+0 = 4 for that column. which some people consider unnatural. 5 + 2. 0 + 6. Adding 4 to 5 then rounds oﬀ the whole sum. starting with the right column and moving to the left. plus 9 extra ones. and since there is no digit in the spot below the number 4. First of all. assume that the digit is 0.3 The Standard Algorithms
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of 10: the sack of 34 potatoes is put in 3 bags of 10 each plus 4 stray ones. Therefore an intelligent way to count the total number of potatoes is to ﬁrst count the total number of bags of 10’s (3 + 2 is 5. we shall discuss why the algorithm moves from right to left. and the 2 in 25 signals that there are 2 tens. and so there are 9 extra). the main emphasis is usually not on simple cases such as 4502 + 263 where the sum of
. In due course. in case no digit appears in a certain spot. which puts the total at 59. It says: The sum of two whole numbers can be computed by lining them up digit-by-digit. Schematically then for 4502 + 263:
+
4 5 0 2 2 6 3 4 7 6 5
(12)
Thus starting from the ones digit (right column). we have: 2+3 = 5. Adding 2 and 3 to get 5.

There is a good reason for this decision: it is in the simple case that one gets to see with unobstructed clarity the main line of the logical reasoning. the addition fact in (14) actually stands for + 5 0 0 2 0 0 7 0 0
because the 5 in 4502 stands for 500 and the 2 in 263 stands for 200. consider 865 + 32: 8 6 5 + 3 2 (13) 8 9 7 Notice that the third column from the right of (12) and the rightmost column of (13) are identical: 5 + 2 (14) 7 Yet. Would this digit-by-digit feature of the algorithm corrupt students’ understanding of place value? Not if students are made to understand that. In this monograph. far from a defect. By contrast. and when that is well understood. the algorithm would lose much of its simplicity: imagine that for the ones place you do one thing. we know that in the context of (12). To underscore the fact that each step of the algorithm is strictly limited to the consideration of a single digit without regard to its place value. this digit-by-digit feature is a virtue for the purpose of easy computation. the more complicated case — which is nothing but the simple case embellished with a particular technique — tends to follow easily. we reverse the emphasis by spending more time on the simple case before dealing with the more complicated case. the addition fact (14) is carried out in exactly the same way in (12) or (13). however. without regard to this diﬀerence. for the tens
. in terms of place value. As far as the algorithm is concerned. but on cases such as 69 + 73 where the process of “carrying” to the next column takes place.3 The Standard Algorithms
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the digits in each column remains a single-digit number. the addition fact in (14) is literally true: it is just 2 + 5 = 7. Had the procedure of the algorithm treated each digit diﬀerently according to its place value. in the context of (13).

where we have made use of the distributive law (10) three times in succession.) Recall an earlier comment made also in §1 about the occasional advantage of explicitly writing down all the powers of 10 in the expanded form of a number.
. recall: 865 = 8 × 102 + 6 × 101 + 5 × 100 32 = 0 × 102 + 3 × 101 + 2 × 100 (15)
(Notice that we have made use of the trivial fact 32 = 032 mentioned at the end of §1. (16)
which is seen to be a parallel description of the addition algorithm in (13). In particular. students should at the same time achieve the understanding that the algorithm is correct precisely because of place value considerations. We now see that the addition algorithm is the method of adding two numbers by adding the digits corresponding to the same power of 10 when the two numbers are written out in their expanded forms. The reasoning is valid in general and is by no means restricted to this special case of 865 + 32. we get 865 + 32 = (8 + 0) × 102 + (6 + 3) × 101 + (5 + 2) × 100 (= 8 × 102 + 9 × 101 + 7 × 100 = 897 ) . The left sides add up to 865 + 32. of course. we see why the algorithm calls for replacing the empty spot under 8 with 0: it is none other than an abbreviated statement of 8 + 0 = 8 in (16). To see this.3 The Standard Algorithms
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place you do another. etc. How eﬃcient can the algorithm be in that case? Moreover. What about the right sides? We can add “vertically”: 8 × 102 + 0 × 102 = (8 + 0) × 102 (= 8 × 102 ) 6 × 101 + 3 × 101 = (6 + 3) × 101 (= 9 × 101 ) 5 × 100 + 2 × 100 = (5 + 2) × 100 (= 7 × 100 ) . We add these two equations. and for the hundreds place you do yet another. in learning the procedural aspect of the algorithm. (It may be instructive for you to read the exhortation near the end of §2 about the need to know that M N + M L = M (N + L). Because the left side of the sum of equations equals the right side of this sum. We will see how this more clumsy notation lends conceptual clarity to what we have to do.

6 ten-dollar bills. thus $865 altogether. such explanations tend to be somewhat tedious and students might lose interest and. This is because. Pedagogical Comments: Should one emphasize discussion of the above type in an elementary school classroom? Few questions in education can be answered with absolute certainty. Later she acquires another stack of bills consisting of
. but there are at least two reasons why an elaborate discussion of the associative law and commutative law in. 865 + 32 = {8 × 102 + 0 × 102 } + {6 × 101 + 3 × 101 }+ {5 × 100 + 2 × 100 } It is at this point that we can apply the distributive law (10) to conclude (16). it is possible to give a naive explanation of the addition algorithm in terms of money. and 5 one-dollar bills.3 The Standard Algorithms
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Activity: Give a similar explanation of (12). on the other. As a teacher. intellectual honesty demands it and. on the one hand. Imagine that someone has 8 hundred-dollar bills.
For the case of 865 + 32. It is important to realize. such details might obscure the main thrust of the argument which is encapsulated in (16). you owe it to yourself to understand this kind of details.
The preceding explanation of the addition algorithm — by this we mean the main ideas but not necessarily the precise notational formalism — would be adequate in most classroom situations. second. we can ignore the braces and rearrange the order of summation of these six terms. From (15). that there are subtle gaps in the reasoning above. however. however. we have: 865 + 32 = {8 × 102 + 6 × 101 + 5 × 100 }+ {0 × 102 + 3 × 101 + 2 × 100 } As noted in connection with the associative and commutative laws of addition (5) and (6). K–5 might interfere with a good mathematics education. Thus. End of Pedagogical Comments. say. so that in the interest of a complete understanding. Thus what we said above about “adding vertically” in (15) is in fact an application in disguise of the associative and commutative laws of addition. we proceed to ﬁll in these gaps. you must be prepared in case a precocious youngster presses you for the complete explanation. First.

On a deeper level. it is most likely the case that you have also found a purely mathematical explanation along the line of (16). but relatively easy to do so by the mathematical method above. she decides on the following strategy: collect all the hundred-dollar bills. the question must be raised as to why it is not suﬃcient to understand the addition algorithm in terms of money alone. it would be extremely clumsy to explain the addition 50. because the algorithm is about numbers and not speciﬁcally about money. we have just seen how the mathematical explanation brings out issues that are hidden in the explanation using money. such as place value and the basic laws of operations of §2 . The mathematical explanation brings a deeper understanding not only of the algorithm itself but also of these related issues. as we shall see. we want an explanation that is suﬃciently robust to be applicable to all numbers big or small. or 865 stars + 32 stars = 897 stars ? By the time you have found the answer. 060. The most superﬁcial answer is that. as we begin to see the interconnectedness of these seemingly disparate concepts. She ﬁnds that she now has 8 (= 8 + 0) 9 (= 6 + 3) 7 (= 5 + 2) hundred dollar bills ten-dollar bills one-dollar bills
So she has $897. why does this explanation also explain the fact that 865 crabs + 32 crabs = 897 crabs. and why we must go through the previous mathematical explanation. In this regard. and 2 one-dollar bills. this issue is no more than a minor reﬁnement of all
. and ﬁnally collect all the one-dollar bills. Before proceeding further. thus another $32. 040 in terms of money. Let us now take up the issue of “carrying” from one column to the next. Exactly as in (13). For example. then collect all the ten-dollar bills. To ﬁnd out how much money she has altogether. Moreover. we should be able to oﬀer an explanation that is valid in all contexts besides money. From our point of view.3 The Standard Algorithms
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3 ten-dollar bills. 001 + 870.

Consider 68 + 59: 6 8 5 9 +
1 1
(17)
1 2 7 The diﬀerence from the previous situations is that. is no longer a single digit number but is 17. In like manner then. i. the sum of digits. the “1” of 12 is carried to the (invisible) hundreds column. but in some other conventions. which is again not a single digit number. and one-dollar bills. The basic explanation is the same as before.3 The Standard Algorithms
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that has been said before. and one-dollar bills. this 1 is taken into account and result in the sum 6 + 5 + 1 = 12.. in the right column. Thus you have 11 (= 6 + 5) 17 (= 8 + 9) ten-dollar bills. such minor notational diﬀerences are irrelevant to the mathematics under discussion. Let us begin with an explanation in terms of money. The algorithm calls for carrying the tens digit “1” of 17 to the next column on the left.
You decide to count the two stacks by counting the ten-dollar bills and the one-dollar bills separately. the 1 is entered above 6 instead. and one-dollar bills. This is indicated by the small 1 under 5. (18)
. and one-dollar bills.e. 8 + 9. Because there is no other number in the hundreds column. Imagine that you have a stack of bills consisting of 6 8 ten-dollar bills. Then in adding the numbers in the tens column.
But 17 one-dollar bills is the same as 1 ten-dollar bill plus 7 one-dollar bills. so it suﬃces to consider only the new features here. the last 1 is recorded in the hundreds column and we get 127 as the ﬁnal sum. So you trade 10 of your one-dollar bills for a ten-dollar bill and you now have 12 (= 11 + 1) 7 ten-dollar bills.
and another stack consisting of 5 9 ten-dollar bills.

thereby getting  1 hundred-dollar bill. This line of reasoning not only arrives at the same answer.  2 ten-dollar bills. commutative. We preface the explanation with the comment that. we will not overwhelm you with such details from now on but will instead ask you to be aware of their implicit presence at every turn. and distributive laws. This is the right time and place to tie up a loose end. which is 127. So 68 + 59 = 127. 68 + 59 = (6 × 101 + 8 × 100 ) + (5 × 101 + 9 × 100 ) = (6 × 101 + 5 × 101 ) + (8 × 100 + 9 × 100 ) Applying the distributive law. while (19) corresponds to carrying the 1 to the hundreds place in (17). Then the ﬁrst step would be: + 6 8 5 9 1 1
. but also exhibits the fact that (18) corresponds exactly to carrying the 1 to the tens place in (17). With this understood. we get 68 + 59 = = = = (6 + 5) × 101 + (8 + 9) × 100 (10 + 1) × 101 + (10 + 7) × 100 (1 × 102 ) + (1 × 101 ) + (1 × 101 ) + (7 × 100 ) (1 × 102 ) + ([1 + 1] × 101 ) + (7 × 100 ). Consider the preceding problem of 68 + 59. having given the detailed reasoning in supported of the computation (13) citing the associative. we give a purely mathematical explanation of (17).
the last line explicitly shows how the 1 is carried to the next column to the left. The reason is one of economy of means : we want to simplify the algorithm to the utmost. Suppose we want to embark on a diﬀerent addition algorithm by starting instead with the leftmost column and move to the right.3 The Standard Algorithms
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But we can also trade in 10 of the 12 ten-dollar bills for 1 hundred dollar bill. and (19)  7 one-dollar bills. twice. It was mentioned after (12) that there is a reason for the algorithm to move from right to left. Finally. in (16).

The fact is that children learn to do many things that they consider unnatural.3 The Standard Algorithms
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Here we have carried the 1 to the hundreds column. we add the right column and get 17. Nevertheless. to discuss this and other whole number algorithms using abstract notation. and this is the decisive factor. we hope that the generality of the underlying mathematical reasoning
. whereas the carrying in (17) is recorded under 59. Brushing teeth is one of them. Yet. All the more so when the addition involves numbers with many digits. A passing comment should be made about the “unnaturalness” of making children do things from right to left. notationwise. This can be recorded in the next row again in accordance with the place value of 17 thus: + 6 8 5 9 1 1 1 7
Then we add these two numbers and arrive at 127 as before: + 6 5 1 1 1 1 2 8 9 7 7
The ﬁrst comment is that this algorithm is certainly correct. children all over the world welcome video or computer games with open arms. One must put things in perspective. the only diﬀerence being that. (17) is the algorithm of choice for most people. In the next step. Moreover. Moreover. the proper execution of this algorithm requires the same technique of carrying the 1 to the next column. but once familiarity sets in. Are we then to believe that they cannot learn to calculate from right to left even after we explain to them why it is important that they do? We have thus far discussed the addition algorithm in terms of concrete numbers for the simple reason that it is very diﬃcult. Comparing the last computation with (17) reveals one virtue of (17): it is compact. here it is done by using two extra rows. A beginner might welcome the left-to-right algorithm at the beginning. and staying neat and clean is another. inducing violent actions on a video screen by manipulating a joystick is quite possibly one of the most unnatural things imaginable for children.

and give an explanation of why the computation is correct. 000. according to the interpretation of the sum of two whole numbers as the length of a concatenated segment. (n − m) + m = n (21)
So again. It also means that in order to check whether a number x is equal to n − m. it suﬃces to verify a statement about addition. 000 − 20.2
A subtraction algorithm
Subtraction has to be understood in terms of addition. and m < n.
3. when added to 19. Thus 37 − 19 is by deﬁnition the number so that. So adding 19 undoes the eﬀect of subtracting 19. to verify a statement about subtraction. 000 − 500.5 Compute 7826 + 7826 + 7826 by the addition algorithm. n]. In particular. Note that m and n are the lengths of the segments [0. 000. it is the same as checking whether x + m = n. This simple fact will be seen to be extremely useful. 000. and give an explanation of why the computation is correct. the preceding equality implies that n − m is exactly the length of the segment from m to n:
. adding m undoes the eﬀect of subtracting m. Similarly.6 Compute 670309 + 95000874 by the addition algorithm. when added to m . and give an explanation of why the computation is correct.4 Compute 123 + 69 + 528 + 4 by the addition algorithm. Rewrite (21) as m + (n − m) = n (the commutativity of addition!). Exercise 3. it is common to think of 37 − 19 as “taking 19 objects from 37 of them”. Activity: 1200 − 500 = ? 580. 000 = ? 580. Thus. Thus: (37 − 19) + 19 = 37. yields n. yields 37. then n − m is by deﬁnition the number so that. if m and n are whole numbers. Therefore. n − m has the following geometric interpretation. Exercise 3. m] and [0. On an intuitive level. 000 = ? 15 × 106 − 7 × 106 = ? In terms of the number line.3 The Standard Algorithms
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Exercise 3. and n − m as “taking m objects from n of them”. 000.

where the simplicity refers to the fact that each of the digits in the ﬁrst number is at least as big as the corresponding digit in the second number. m. As before. n − m] and [0. a. we begin with the simple case. the purpose of the standard subtraction algorithm is to relieve the tedium of counting backwards 257 times from 658 in order to compute 658 − 257. 5 − 5 = 0. the following subtraction facts are useful.g.10 Schematically: − 6 5 8 2 5 7 4 0 1 (22)
For the explanation of the algorithm.3 The Standard Algorithms
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0
m
'
n n−m
E
We can also look at (21) as is: (n − m) + m = n. The mathematics underlying this algorithm (to be introduced presently) is so similar to that of the addition algorithm that we can aﬀord to be brief. m > b and n > c. 658 − 257. This means that we can arrive at the point n − m by going from n to the left for a length of m:
0
n−m
'
n m
As is the case with the addition algorithm. and 8 − 7. the subtraction of 658−257 is reduced to three single-digit computations: 6−2.
. 5 − 5. e. b. The algorithm calls for lining up the digits of the two numbers column-by-column from the right (as in the addition algorithm) and then do subtraction of single digit numbers in each column. This says that the concatenation of [0.. m] is a segment of the same length as [0. and c are any whole numbers so that l − a. n. starting with the right column and move left: 8 − 7 = 1. Suppose l. 6 − 2 = 4. n]. we have: (l + m + n) − (a + b + c) = (l − a) + (m − b) + (n − c) m(l − a) = ml − ma
10
(23) (24)
Again.

having l. We have thus completed the formal proofs of (23) and (24)
. First note that (23) makes sense.e. To see why.3 The Standard Algorithms
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Equation (23) is entirely plausible if it is interpreted in terms of concrete objects. then taking a + b + c oranges away from the three piles combined would leave behind the same total number of oranges as taking successively a oranges from the pile of l oranges. m and n oranges in each pile. Equation (24) is a variant of the distributive law (10) and can be made believable using oranges in exactly the same way. which can be further simpliﬁed by use of x + a = (l − a) + a y + b = (m − b) + b z + c = (n − c) + c = l = m = n
Thus the left side of (♣) is l + m + n. Then (24) becomes the statement that mx = ml − ma. and we want to prove that (l + m + n) − (a + b + c) = x + y + z.
It is also instructive to directly prove (23) and (24). For example. then the right side of (23) becomes x + y + z. Simlarly. m > b. b oranges from the pile of m oranges. and n > c to conclude that (l + m + n) > (a + b + c). let x = l − a. This shows that (♠) is also true. Again by the remark after (21). let x = l − a. to show that (24) is true. y = m − b. i. because we will get to know the deeper meaning of the deﬁnition (21) in the process. the left side of (♠) is m(x + a) which in virtue of x = l − a is equal to m([l − a] + a) = ml. Now to prove (23). this is the same as checking ((a + b + c) + (x + y + z) = (l + m + n) (♣)
By the general associative law and general commutative law of addition. if you have three piles of oranges. respectively. it is in fact true that (l + m + n) > (a + b + c) so that the left side of (23) makes sense. and c from the pile of n oranges. this would be true if we can show mx + ma = ml (♠)
But by the distributive law. and z = n − c.. which shows that (♣) is true. the left side of (♣) is equal to (x + a) + (y + b) + (z + c). we make repeated use of the third assertion of (11) near the end of §2 (to the eﬀect that A > B and C > D imply A + C > B + D) and use the assumption of l > a. According to the remark after deﬁnition (21).

cannot be carried out using whole numbers. Schematically:
4 9 9
5 − 4
0 0 3 4 6 5 5 3 8
. we can take 1 to bring it down to the tens digit. From this 10. So we go to the hundreds digit and hope to take 1 from there. which is unfortunately zero. Just to be sure that this message gets across. so long as a is not smaller than b. We are now back to our original problem of getting 1 from the tens digit to change the 3 to 10 + 3 = 13 . and the tens digit becomes 9. and that in the hundreds digit becomes 9 − 4 = 5 . the hundreds digit becomes 9 (instead of 10). So now. and the tens digit is now 10+0 = 10 .3 The Standard Algorithms
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because the more complicated notation would have obscured the underlying reasoning. it is nevertheless worthwhile to point out the critical role played by the associative law of addition in the subtraction algorithm. We add the usual remark that the mathematical reasoning behind the preceding explanation is perfectly general and is applicable to any subtraction a − b for any whole numbers a and b. The subtraction in the tens column becomes 9 − 6 = 3 . Although it is not a good policy in general to over-emphasize the role of the laws of operations discussed in §2. We try to take a 1 from the tens digit of 5003. the ﬁrst subtraction in the right column. So 5 becomes 4 in the thousands digit of 5003. We do so. Again it is 0. but the hundreds digit of 5003 becomes 10 + 0 = 10 . in the process. The thousands digit is of course 4. As before. This then requires going all the way to the thousands digit “5” of 5003 to take 1 from 5. 3 − 5. we will work out another example: 5003 − 465. the subtraction in the ones column becomes 13 − 5 = 8 .

This applies in particular to the subtraction algorithm. we have: 5003 − 465 = = = = (4000 + 900 + 90 + 13) − (0 + 400 + 60 + 5) (4000 − 0) + (900 − 400) + (90 − 60) + (13 − 5) 4000 + 500 + 30 + 8 4538
We note that the subtraction algorithm is again one that works from right to left. we will dispense with the use of parentheses altogether): 5003 = = = = = 5000 + 3 4000 + 1000 + 3 4000 + 900 + 100 + 3 4000 + 900 + 90 + 10 + 3 4000 + 900 + 90 + 13
So using the generalized version of (23) for four pairs of numbers. 30024 − 8697 = 25 + 29999 − 8697 = 25 + 21302 = 21327. Similarly. Let us give one such example: the preceding problem 5003 − 465 can be done very simply as follows: 5003 − 465 = 4 + 4999 − 465 = 4 + 4534 = 4538 . Activity: Do the preceding subtraction 5003 − 465 from left to right. Just as in the case of the addition algorithm. there are usually tricks to make computations with them much more pleasant. one can work from left to right. It is worth repeating that there is absolutely nothing unnatural about teaching children to do something from right to left.
.
For special numbers.3 The Standard Algorithms
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The explanation using the associative law of addition follows (this time. but the amount of backtracking needed for making corrections is even greater here than in the case of addition. and compare with the computation from right to left.

the algorithm says:
. for instance.
The preceding algorithm for subtraction when the ﬁrst number is close to a multiple of 10n is so striking that it would have spoiled the fun if we had called attention to a technical point. from left to right or right to left.
As a ﬁnal note on the subtraction algorithm.g. Then by changing the ﬁrst number to a small number plus another one with a row of 9’s. The point is that the step 4 + 4999 − 465 = 4 + 4534 actually involves the associative law for integers (positive and negative whole numbers. mathematics must be served. So: − 7 5 6 3 8 9 4 [[−3]] [[−3]]
where each [[−3]] indicates the result of the column subtraction in that column. e. where n is any whole number. see Chapter 5). To illustrate. 5003 − 465. but now we can use negative numbers to record the diﬀerences and we are free to do the column-by-column subtractions in any order. we have (4 + 4999) − 465 = (4 + 4999) + (−465) = 4 + (4999 + (−465)) = 4 + 4534 The reason why the preceding is worthy of a separate discussion is of course the fact that the associative law in its naive form is not always valid for subtraction. The mystery surrounding the associative law for subtraction would be dispelled once we do integer operations correctly. consider the previous problem of 756 − 389 : We continue to do column-by-column single digit subtractions. (19 − 4) − 5 = 19 − (4 − 5). One can even do it from left to right if one wishes.. Precisely. The same trick can be used for any subtraction problem in which the ﬁrst number is close to a multiple of 10n . See Chapter 5. Take. Nevertheless. the subtraction can be done with no eﬀort.3 The Standard Algorithms
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The point here is that the subtractions with rows of 9’s in the ﬁrst number can be done without trading and can therefore be done easily by mental math. To get the ﬁnal answer. if only discretely this time. so we attend to this technical point here. there is a natural alternative algorithm if negative numbers — together with a few simple arithmetic properties — are allowed to be used.

12 Compute 800.7 Give an interpretation of (22) in terms of money. the second algorithm should be a viable option in schools. Exercise 3. b. once using the standard algorithm.13 Compute 26. but the second may be less prone to computational errors. at least if the user is ﬂuent with negative numbers. (b) Suppose c < a and c < b. Exercise 3.14 Let a. [[−3]]. (a) Prove that a + b < c is the same as a < c − b. Exercise 3. with or without money.3 The Standard Algorithms
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treat 4[[−3]][[−3]] as if it were a whole number with digits 4. 982 in two diﬀerent ways. 400 − 770. Because teaching third graders about the most elementary aspects of negative numbers is a realistic goal. c be whole numbers. Prove that a < b is the same as a − c < b − c. 004 − 8325 two ways. (b) Do the same with 76431 − 58914. and [[−3]]. Notice that the subtractions here are much more tractable than those in the original. Exercise 3. Exercise 3.8 Explain to a fourth grader why the subtraction algorithm for 563 − 241 is correct. Thus: 4[[−3]][[−3]] = (4 × 102 ) + ((−3) × 101 ) + ((−3) × 100 ) = 400 − 30 − 3 = 370 − 3 = 367 . Exercise 3.
. with and without the use of money.11 (a) Use the subtraction algorithm to compute 2403 − 876 and explain why it is correct. Exercise 3. and write it out in expanded form. and once using the preceding “negative number” algorithm. The explanation is simple enough and is based on (23) and (24): 756 − 389 = {(7 × 102 ) + (5 × 101 ) + (6 × 100 )}− {(3 × 102 ) + (8 × 101 ) + (9 × 100 )} = ([7 − 3] × 102 ) + ([5 − 8] × 101 ) + ([6 − 9] × 100 ) = 400 + ((−3) × 10) + (−3) = 367 There are pros and cons regarding which of two algorithms is “better”. Exercise 3. The ﬁrst is simpler.9 Give an interpretation of (25) in terms of money. and explain why what you have done is correct.10 Explain to a fourth grader why the subtraction algorithm for 627 − 488 is correct.

Instead of adding 3 to itself 826 times. Bearing in mind the leitmotif enunciated at the beginning of the section.e. because it lies at the heart of all multiplication problems. we apply the distributive law one more time: 826 × 3 = {(8 × 102 ) + (2 × 10) + 6} × 3 = (8 × 3) × 102 + (2 × 3) × 10 + (6 × 3) (27) (26)
where of course we have also made use of the associative and commutative laws of multiplication to conclude that.g.
. the distributive law does the breaking up: 826 × 73 = 826 × (7 × 10 + 3) = (826 × 7) × 10 + (826 × 3) Thus the multiplication (of 826) by a multi-digit number 73 has been reduced to two simpler computations: multiplication by 7 (i. as soon as we know the products of single-digit numbers: 8×3. we proceed to break up the computation into a series of computations involving one digit at a time. as given by (26). e. from (27) we obtain: 826 × 3 = (24 × 102 ) + (6 × 10) + 18. We now further break up these two tasks into yet simpler tasks. Let us ﬁrst look at 826 × 3. In this case..15 Find shortcuts to compute the following: 8 × 875 = ? 9996 × 25 = ? 103 × 97 = ? 86 × 94 = ?
3. 2×3.3
A multiplication algorithm
We next take up the question of how to compute the product of two numbers such as 826 × 73 without having to add 73 to itself 826 times. In any case.. See the discussion in §2 above equation (10). and 6×3.3 The Standard Algorithms
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Exercise 3. Thus 826 × 3 will be computable according to (27). This is why a ﬂuent knowledge of the multiplication table is essential. (8 × 102 ) × 3 = 8 × 3 × 102 . 826 × 7) and multiplication by 3 (826 × 3).

We now know:
826 × 7 = 5782 826 × 3 = 2478 .. when adding them. 826 × 3 and 826 × 7 — and. Recall that we are looking at the multiplication algorithm with a singledigit multiplier because (26) reduces the general case to this special case. the discussion of the addition algorithm right above (12). for example).
. compute the two products with single digit multiplier — i.e.e.. it is customary to omit the “0” at the end of 57820 and just write: × + 5 6 8 2 7 2 4 7 7 8 2 0 2 9 6 3 8 8
(31)
We are in a position to summarize this algorithm as follows: To compute say 826 × 73. so that schematically we have: × + 5 6 8 2 7 2 4 7 7 8 2 0 2 9 6 3 8 0 8
Because we are used to treating an empty slot as a zero (cf. as in (31). take the digits of the second factor 73 individually. shift the one involving the tens digit (i. say (26). According to (26) then. It is time to return to the general case. 826 × 73 = (5782 × 10) + 2478 = 2478 + 57820 .3 The Standard Algorithms
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and the last line explains the carrying of the 1 to the hundreds column. 7) one digit to the left.

Give a precise explanation of the preceding algo-
Pedagogical Comment: In a classroom.. For example. this means we run the algorithm from left to right. This should be carefully explained to them in terms of place value: we are actually looking at 826×70 and the shifting of digit is caused by the presence of the “0” in the ones digit.3 The Standard Algorithms
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This is commonly called the standard multiplication algorithm. the most salient feature of this algorithm that catches students’ attention may well be the shifting of the product involving the tens digit (i. Mathematically. It is worth noting that even the algorithm with a single-digit multiplier can be carried out from left to right. Other variations are possible.e. ﬁrst multiply by 7 before mutiplying by 3. End of Pedagogical Comment. In order to ensure that the generality of the preceding reasoning behind the algorithm is understood. 6718 × 5 can be done this way: × 6 7 1 8 5 3 0 3 5 5 + 4 0 3 3 5 9 0
Activity: rithm. we do not consider such formal diﬀerences to be a diﬀerence at all. let us brieﬂy explain how to do a more compli-
. one alternative algorithm is × 5 + 6 8 2 7 7 8 2 2 4 7 0 2 9 6 3 8 8
Essentially. 826 × 7) to the left by one digit. For example.

You are however asked to use reasoning to quickly dospatch it by narrowing down the choices. because the 3 of 364 is the hundreds digit. gives a 4-digit number that begins and ends with a 6? (b) List all the 3-digit numbers which have the following properties: the sum of the digits is 12. × 2 3 1 1 5 8 1 9 1 5 3 1 6 1 8 2 7 6 4 0 8 2 2 8
(32)
+ This is because:
527 × 4 = 2108 527 × 6 = 3162 527 × 3 = 1582 . the product 527 × 3 is shifted two digits to the left in keeping with the fact that it is really 527 × 300 that we are adding.16 Explain to a 4th grader why the multiplication algorithm for 86 × 37 is correct.e. (Clearly this problem can be done by guess and check. Exercise 3. 3162. Exercise 3.) Exercise 3.. the ones digit is 5).17 (a) Which 2-digit number. when multiplied by 89.3 The Standard Algorithms
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cated problem.18 Use the multiplication algorithm to compute × 1 8 5 0 0 0 9 2 ?
. and when multiplied by 15 they give a 5-digit number which ends with a 5 (i. and 1581 in (32). and so by the distributive law: 527 × 364 = 527 × (3 × 102 + 6 × 10 + 4) = (527 × 3) × 102 + (527 × 6) × 10 + (527 × 4) = 158100 + 31620 + 2108 The last line explains the vertical alignments of the digits of the three products 2108. In particular.

Before we can come up with an answer. Such vagueness would not serve any purpose because unless the student already knows what “division” means. or if they do. This knowledge is critical not only for the long division algorithm of §3. how do we compute the “quotient” of 6810255956001 ÷ 28747 if we do not know its precise meaning? The usual attempt at an explanation of division would mention taking away multiples of 28747 until “the remainder” is “smaller than 28747”. however.19 Compute 4208 × 879 by the multiplication algorithm and explain why it is correct.5 below but also for the discussions of fractions in Chapter 2 and decimals in Chapter 4.
3. This unfortunately begs the question of what a “quotient” is and whether a negative number may be considered to be “smaller than” 28747. we need to know precisely what the “quotient” of a ÷ b means and what the “remainder” means.4
Division-with-remainder
When asked to divide 23 by 4. quotient The answer in division. but this still leaves out a clear statement about what a “quotient” is and what “left over after dividing” means. the above explanations give no information. we likewise want to know what the quotient and remainder are when a is divided by b.3 The Standard Algorithms
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and explain why it is correct.
. Most school textbooks do not deem it necessary to explain these concepts. If you look more closely at the text proper. they typically say the following: division An operation on two numbers that tells how many groups or how many in each group. In general. you would ﬁnd the statement that the remainder should be less than the “divisor” b. given whole numbers a and b with b = 0 . For example. Compare with the same algorithm applied to × 5 0 0 0 9 2 1 8 ?
Exercise 3. remainder The number that is left over after dividing. we all know the answer: the quotient is 5 and the remainder is 3.

mathematics is characterized by its WYSIWYG quality — what you see is what you get — and you have no need to assume anything that is not already explicitly stated. (In particular. subtract) successive multiples of b from a: a − 0. Division a ÷ b is roughly speaking repeated subtraction. see the discussion in §4 below. Among all branches of knowledge. q is that whole number so that a ≥ qb
11
but
a < (q + 1)b. let us ﬁrst ﬁx the meaning of a ÷ b where a and b are whole numbers and b = 0 . no guesswork is needed for its mastery. we must have x ≥ b).e. . . Symbolically though. division is to multiplication as subtraction is to addition: one undoes the other. In greater detail.) Generally speaking. Recall that multiplication is repeated addition: by deﬁnition. We want to let them know that it is an open book that everybody can read. the product mb is called a multiple of b. there are certain wrinkles to this simple-minded statement. a − 4b. if m is a whole number. and we will be careful to address them. and we intend to continue doing it for the rest of this monograph. the m-th multiple of b.) Intuitively. . However.
. This is what we have been doing so far. by deﬁnition because 0 = 0 × b. . qb = b + b + · · · + b
q
(see (2) of §1). a − 3b. or more precisely. this way of doing things is very awkward. In symbols. until eventually we come to a multiple qb of b so that the next multiple — which is (q + 1)b — exceeds a. a − 2b. First of all. what we are going to do is to repeatedly subtract b from a until we get to the point where the next subtraction will not be possible because what is left is smaller than b (recall: for x − b to make sense.
(33)
An explanation of why it is the same is given in the next ﬁne-print indented passage. we want to convey the clear message to our children that in mathematics. but the precise meaning of this phrase requires a rather long-winded explanation. To return to division. 0 is a multiple of b.3 The Standard Algorithms
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As teachers. a − b. what we do is to take (i.. (We will always assume b = 0 in this situation because we do not want to divide by zero. This is another way of expressing the fact that every conclusion we draw in mathematics depends completely on what is stated explictly up front. so we do something which is the same11 but which is easier to express.

by (33). namely. To complete the terminology. however. if a = 23 and b = 4.2)
So we get a − qb < b. It is intuitively clear. that after we have taken q multiples of b’s from a.e. because we cannot perform the next subtraction a − (q + 1)b for the reason that a < (q + 1)b. If a = qb for some whole number q. 3 × 4 = 12. and the remainder is 23 − (5 × 4) = 3. a − qb < b. when a = qb.14(b)). the quotient of 12 ÷ 3 is 4. and we stop at the 6th multiple of 4 because already 23 ≥ (5 × 4) but 23 < (6 × 4).. We may therefore summarize the preceding two facts in the following double inequality: the remainder a − qb satisﬁes 0 ≤ a − qb < b (34)
Note that the left inequality sign of (34) is a weak inequality (i. then we get the happy coincident that 12 = 4 × 3. We ﬁrst give a numerical proof. For example. then the multiples of 4 are 4. the inequality does not change (see Exercise 3. i. what is left behind is less than b. we say b divides a. In symbols: b|a.3 The Standard Algorithms
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By deﬁnition. so a − qb ≥ 0. Incidentally. and we shall prove precisely below. but presently we will also give a pictorial proof that makes the reasoning perfectly obvious.e. 4 × 4 = 16. Note that “ b divides a” says exactly the same thing as “ a is a multiple of b”. If we take a = 12 and b = 3. 5 × 4 = 20.. 12 is exactly the 4th multiple of 3. By taking qb away from both sides. Here then is the numerical proof: from (33) we have a < (q + 1)b. In this case.g. It follows that the quotient q is the largest multiple of b that one can take away from a.
Therefore the quotient of 23 ÷ 4 is 5. e. We call a − qb the remainder of a ÷ b. with remainder 0. We now proceed to bring out a critical property of the remainder that was alluded to above. we call b the divisor and a the dividend of the division a ÷ b. we also know from (33) that a ≥ qb. allowing for equality) because the remainder does equal 0 sometimes.. 2 × 4 = 8. So we get a − qb < (q + 1)b − qb = {(q + 1) − q}b = b (by (24) of §3. we call this q the quotient of a ÷ b. 6 × 4 = 24.
.

36. 108. 30. the quotient of 97 ÷ 12 is 8. As another example.. 84. let a be between qb and (q + 1)b. b units apart. and the remainder of 97 ÷ 12 is the length of the segment [96. as shown. Consider 27 ÷ 6. 48. 97
c
0
12
24
36
48
60
72
84
96
108
Again. 188 ÷ 37. or right at one of the multiples. 98 ÷ 19. The multiples of 6 are 0. which is clearly less than the length between 24 and 30. Thenquotient is 4 because 24 = 4 × 6. 27 0 6 12 18 24
c
30
The picture clearly displays the fact that 27 is trapped between the two multiples of 6: 24 and 30. 12. 24. 24. then the multiples of b are equally spaced markers (= points) on the number line. 96. 46 ÷ 9. 97] (the segment between 96 and 97). The multiples of 12 are: 0. In general.3 The Standard Algorithms
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Activity: Find the quotient and remainder in each of the following divisions by listing the multiples of the divisor (but do not use “long division”. the dividend 97 is trapped between the two multiples 96 and 108 of 12. The whole number a has to be trapped between two of these multiples. The remainder of 27 ÷ 6 is just the length of the segment between 27 and 4 × 6 = 24. 60. we have a ÷ b. 18. We now give the pictorial representation of the quotient and remainder using the number line. etc. In the former case. 108]). 12. 72. which is less than 6 (= the length of the segment [96. a
c
0
b
2b
3b
···
qb
(q + 1)b
. whatever that means): 33 ÷ 7. Therefore the remainder 27 − (4 × 6) < 6 (= 30 − 24). Because 96 = 8 × 12. 6. consider 97 ÷ 12.

This theorem is more commonly cast in a diﬀerent form. there exist a whole number q. However. for 5 ÷ 32. On the other hand. there exist a whole number q. The condition (34) now says 0 ≤ r < b. i. (q + 1)b] (= b).e. as shown. let it be at qb. Let the remainder a − qb be denoted by r. a
c
0
b
2b
3b
···
(q − 1)b
qb
In this case. (35)
An added remark about the division-with-remainder will be relevant in the discussion of long division (§3. which can be rewritten as a = qb + r. We can summarize our discussion in the following theorem. with quotient 0 and remainder 29. called the remainder of a ÷ b. with b > 0. so that the quotient is 0 and the remainder is 5. and it is clearly smaller than the length of the segment [qb. a = qb and of course the remainder is 0. The point is that the division-with-remainder makes sense for any a ÷ b.
. Ordinarily when one divides a by b.5) and the decimal expansion of a fraction (§4 of Chapter 4). the remainder is just the length of the segment [qb. For example. Given any two whole numbers a and b. a]. 0 ≤ a − qb < b. and a whole number r. as follows. if a is at one of the multiples of b. we have 5 = (0 × 32) + 5. Hence. then a − qb = r by deﬁnition. we have 29 = (0 × 127) + 29. we have an equivalent formulation of Division-with-Remainder: Division-with-Remainder (Second Form). so that the remainder a − qb satisﬁes (34). Division-with-Remainder. the relative sizes of a and b are irrelevant. with b > 0. there is an implicit assumption that a is bigger than b. called the quotient of a ÷ b. Given any two whole numbers a and b. in the above statement of division-with-remainder. For 29 ÷ 127.3 The Standard Algorithms
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Then.. so long as b > 0. so that a = qb + r where r satisﬁes 0 ≤ r < b. called the quotient of a ÷ b.

it is also important to realize that intuition need not be the sole arbiter of mathematical truths. what is left behind should be the same. observe again as a consequence of the deﬁnition of subtraction in (21) that x = (a − M ) − N means x + N = a − M. ((a − b) − b). . and in general (· · · ((a − b) − b) · · · − b) = a − qb
q
(†)
for all whole numbers q. a − 3b. in the above. (((a − b) − b) − b). . Then (‡) becomes the statement that x = a − (M + N ). However.2 — it suﬃces to prove x + (M + N ) = a. q = 2. We can now prove (†) in succession for q = 1. because whether one takes b away from a one at a time. ( ) So we have: x + (M + N ) = = = = x + (N + M ) (x + N ) + M (a − M ) + M a (commutative law of +) (associative law of +) (by ( )) (by deﬁnition of subtraction in (21))
This proves x + (M + N ) = a. and therewith (‡). B are whole numbers so that (a − M ) − N ≥ 0. Then we must prove: ([a − b] − b) − b = a − 3b. . We used instead a − b. ( )
. . . The second is the the omission of the uniqueness of the quotient in the division-with-remainder. then (a − M ) − N = a − (M + N ) (‡)
Here is the reason: Let x = (a − M ) − N . a − 2b. We begin with an observation: if A. etc. To this end. Let q = 2. Next let q = 3. . ( )
but this follows directly from (‡) by letting M = N = b. (((a − b) − b) − b) = a − 3b. What needs to be pointed out is that ((a − b) − b) = a − 2b. The ﬁrst is the explanation of why “repeated subtraction of b from a” is the same as ”subtracting successive multiples of b from a”. then (†) states that [a − b] − b = a − 2b. so we shall give a precise proof of (†) which also happens to be instructive. — we recall the remark made after (21) in §3. . we only need to prove x = a − (M + N ). Of course (†) is intuitively obvious. to prove (‡). . If q = 1. Repeated subtractions of b from a means of course (a − b).3 The Standard Algorithms
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We now tie up two loose ends left dangling in the preceding discussion. or take qb away from a all at once. and for this. . . Therefore. (†) merely says a − b = a − b. . q times in succession. q = 3. .

The next thing to prove is the case of (†) for q = 5. with the pattern of proof ﬁrmly established. See the picture:
0
m
n
7
We now apply this simple observation to 31÷7 by letting m = 31−(4×7) and n = 31−(s×7). Then we have to prove (([a − b] − b) − b) − b = a − 4b. n < 7. As before. we have: m−n = {31 − (4 × 7)} − {31 − (s × 7)} = (31 − 31) + {(s × 7) − (4 × 7)} = (s − 4) × 7
Because we know ahead of time that m − n is a whole number. What this means is that suppose we have 31 ÷ 7. we may assume m ≤ n. — and we may as well assume m ≤ n — then it is quite clear that the diﬀerence n − m is a whole number which is not a nonzero multiple of 7.3 The Standard Algorithms
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This is true because: ([a − b] − b) − b = (a − 2b) − b = a − (2b + b) = a − 3b. Then we know that m−n is not a nonzero multiple of 7. it has to be a multiple of
. Using the simple fact that (l − a) − (m − b) = (l − m) + (b − a). We next turn to the division-with-remainder and point out that the quotient q of the theorem is actually unique. we see that (s − 4) × 7 must be a whole number. to be proved in Chapter 5. The reason for this is based on a general fact: if we have two whole numbers m and n so that 0 ≤ m. etc. and 0 ≤ 31 − (4 × 7) < 7. The theorem implies that this s must be equal to 4 too. (by ( )) (by (‡) with M = 3b and n = b)
as desired. Therefore (†) is true for all whole numbers q. that rest of the argument would proceed would proceed in similar fashion. Now suppose there is another whole number s so that 0 ≤ 31 − (s × 7) < 7. (by ( )) (by (‡) with M = 2b and n = b)
Now let q = 4. and this is true because (([a − b] − b) − b) − b = (a − 3b) − b = a − (3b + b) = a − 4b. Then we know the quotient is 4. and as such. But it is clear by now.

It follows that if a = qb.
. there is no harm 1 in using them for illustration. respectively. the ﬁrst example 23 ÷ 4 of this subsection has (as we know) quotient and remainder equal to 5 and 3.. the quotient q of a÷b is unique in the sense that if s is another whole number so that also 0 ≤ a − sb < b. if the quotient is unique. and we proceed to discuss it in some detail. where q is the quotient. s = 4. So let a = qb. the uniqueness is important because we talk freely about the quotient of a division and the remainder of a division.g. in which case s − 4 = 0. The requirement that both q and r be whole numbers is critical to their uniqueness. it is customary to write q = a ÷ b. so it can only be the zero multiple. So without being aware of it. then by the deﬁnition above. 7 and 4 are repalced by a. In general. we can write many such equations at will. which is another way of writing a = (ab) ÷ b. such subtlety is usually glossed over. however.
The special case of division-with-remainder where the remainder is 0 occupies a place of distinction. for example. then necessarily s = q. Exactly as claimed. 23 =
22 4
×4+ 1
and
0≤ 1 <4
This should serve as reminder how delicate the uniqueness of the quotient and remainder really is. then we could have 23 = 5 +
1 4
×4+ 2
and
0≤ 2 <4
(Although we have not yet taken up the subject of fractions. In school mathematics. Conceptually. e. While one cannot say that such negligence does elementary school mathematics great harm. if we are allowed to use fractions. However. The proof is identical to the preceding argument as soon as the numbers 31. and q.) This would give a “quotient” of 5 + 4 and a “remainder” 2. In this case. In fact. For example. More is true. then (a ÷ b) × b = a and (ab) ÷ b = a (36)
(It is of some value to point out that the second assertion is strictly a consequence of the deﬁnition: let x = ab. We already know that it cannot be a nonzero multiple of 7. we hope nevertheless to have convinced you that. b. which is to say. as a teacher.3 The Standard Algorithms
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7. then so is the remainder a − qb.) The equations in (36) clearly display division and multiplication as two operations that undoes each other. respectively. Moreover. we tacitly assume the uniquness in question. you should be aware of it. We will not repeat the argument. a = x ÷ b.

3 is just
. We know that 15 can be partitioned into 3 groups of 5’s. Thus if we divide 15 (= a) objects into 5 (= b) equal groups. Suppose we divide the a objects into b equal groups. how many are in each group? Let a = 15 and b = 5 so that 3 = 15 ÷ 5. Activity: follows: In a third grade textbook. there will be 3 objects in each group. However. there is another common way in which we use division. Although you can see easily in this special case that the answer to both problems is 6. we have a ÷ b is the total number of groups when a objects are partitioned into equal groups of b objects This is called the measurement interpretation of division (in case there is no remainder). we get 5 groups of 3’s and 15 = 3 + 3 + 3 + 3 + 3 = 5 × 3.3 The Standard Algorithms
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If q = a ÷ b. You can divide to ﬁnd how many to put into each group. Recalling once again that q = a ÷ b. with 6 counters in each group. Suppose as usual we have a objects and q = a ÷ b. and 6 by 27 in the above problem. (A) Suppose you have 18 counters and you want to make 6 equal groups. and we represent it pictorially by dots as follows: • • • • • • • • • • • • • • • Now if we count the dots by columns. discuss which of these two divisions uses the measurement interpretation and which requires a new understanding of division. we would exhaust a. It may help you to think more clearly if we replace 18 by 4023. division is introduced as
You can use counters to show two ways to think about dividing. But of course. the fact that a = qb = b + b + · · · + b (q times) means that if we take b objects from a each time and do it q times. (B) Suppose you have 18 counters and you want to put them into equal groups.

Drastic oversimpliﬁcations are involved. you are making a measurement division of the amount of ﬂuid in your punch bowl by the amount of ﬂuid in your cup. one rarely manages to drive at a constant speed for more than a few minutes in real life. Example There is no better illustration of the two meanings of division than the problem of traveling (or motion). Comparing with 15 = 3 + 3 + 3 + 3 + 3 = 5 × 3.. then a = qb. yields the same number as the measurement interpretation) is due to the fact that multiplication among whole numbers is commutative. If you want to ﬁnd out how many cups of punch there are in the bowl. you will ﬁnd both kinds of division done around you. In general then.e. (Warning to the reader: while this may seem like a good example of contextual learning. The fact that this interpretation is valid (i. we immediately recognized that the commutativity of multiplication 3 × 5 = 5 × 3 is at work. Suppose a car goes from town A to town B at a constant speed. both meanings of division are common place in everyday life. For example. in the preceding Activity. which means that the distance traveled within any one-hour time interval is always a ﬁxed constant. For instance. then a ÷ b is also the number of objects in each group when a objects are divided into b equal groups. so that if we divide a objects into b equal groups. with a objects and q = a ÷ b. we should not delude ourselves into believing that this is anywhere close to a “real world” situation. By the commutativity of multiplication. On the other hand. This is called the partitive interpretation of division (in case there is no remainder). a = qb = bq = q + q + · · · + q (b times). Suppose you give a party and you make a bowl of punch. This leads to the following: assuming q = a ÷ b. There is also the implicit idealization in that the car is driving on a freeway that connects the two towns in a straight-line.3 The Standard Algorithms
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the quotient of 15 ÷ 5. task (A) requires that you use the partitive interpretation of division while task (b) requires the measurement interpretation. At the risk of pointing out the obvious. there will be q objects in each group. How often does this happen in everyday driving?) A typical question is then the following:
. Everywhere you look. if four people decide to drink up the whole bowl of punch but wish to exercise caution by ﬁrst computing how much ﬂuid each must be prepared to take in if each drinks an equal amount. then these people will be doing partitive division of the amount of ﬂuid in your punch bowl by 4. which by deﬁnition is equal to 3 × 5 (= qb.

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suppose the distance between towns A and B is 264 miles, and the car gets to town B after 4 hours, what is the speed? If 264 miles is covered in 4 hours, we just partition 264 into 4 equal parts and the number of miles in one part is the number of miles traveled n one hour (“equal parts” because speed is assumed constant). This is the partitive meaning of the division 264 , which is 66. So the speed is 66 miles per hour. Another typical question 4 is the following: suppose the speed is a constant 58 miles per hour and the distance between A and B is 522 miles, how many hours does it take to go from A to B? Here, we know that the car would be 58 miles from A after 1 hour, 58 + 58 = 116 miles from A after 2 hours, 58 + 58 + 58 = 174 miles from A after 3 hours, etc. So the question becomes how many 58’s there are in 522. This is the measurement interpretation of 522 , which is 9. So it 58 takes 9 hours to go from A to B. To summarize: For motion in constant speed, computing the speed when distance and time are given is a partitive division problem, while computing the time to travel a certain distance at a given speed is a measurement division problem. Finally, we give another geometric interpretation of division without remainder. In §2, we introduced an area model for multiplication. According to this model, 2 × 3, for example, would be modeled as the area of the rectangle with vertical side equal to 2 and horizontal side equal to 3:

Now suppose we ask for 6 ÷ 3 = ? From the point of view of the area model, this means we have a rectangle with area equal to 6 and a horizontal side equal to 3, and we want to know what the length of the vertical side is:

?

area = 6 3

'

E

3 The Standard Algorithms

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Similarly, the division 35 ÷ 7 may be interpreted as asking for the length of the vertical side of a rectangle with area 35 and with the horizontal side equal to 7:

?

area = 35

'

7

E

Of course, for whole numbers, such a geometric interpretation of division is no more than slightly entertaining. However, when we come to the division of fractions, the geometric interpretation would acquire added signiﬁcance. Exercise 3.20 Is 24 the quotient of 687÷27? Is 13 the quotient of 944÷46 ? Explain. (No calculator allowed.) Exercise 3.21 Is 6977 the remainder of 124968752 ÷ 6843 ? Why? (No calculator is allowed.) Exercise 3.22 . By taking multiples of the divisor, ﬁnd the quotient and remainder in each of the following cases: 964 ÷ 31, 517 ÷ 19, 6854 ÷ 731, 4972086 ÷ 873, and 4972086 ÷ 659437. Pedagogical Comment: The use of (only) a four-function calculator for the last two items involving 4972086 is allowed. However, it would be instructive to ﬁrst ask for a ballpark ﬁgure of the quotient without the use of a calculator. This is a good exercise in making estimates, and would save a lot of guess-and-check in getting the correct quotient in each case. It should also be mentioned that there is an eﬀective way to use the (four-function) calculator to get the quotient and remainder without any trial and error. How to do this and why it is true should lead to an interesting classroom discussion (one that in fact presupposes some knowledge of decimals). End of Pedagogical Comment.

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Exercise 3.23 Let r be the remainder of a ÷ b. Suppose a = mA and b = mB for some whole numbers m, A and B. Let R be the remainder of A ÷ B. What is the relationship between R and r? Give a detailed explanation of your answer. (Caution: This problem is deceptive because it seems almost trivial, but the explanation is actually quite subtle and it requires the use of the uniquness of both the quotient and remainder which is discussed in the indented ﬁne-print passage of this subsection.) Exercise 3.24 You give your ﬁfth grade class a problem: A faucet ﬁlls a bucket with water in 30 seconds, and the capacity of the bucket is 12 gallons. How long would it take the same faucet to ﬁll a vat with a capacity of 66 gallons? How would you explain to your class how to do this problem? Exercise 3.25 Consider the following two problems: (a) If you try to put 234 gallons of liquid into 9 vats, with an equal amount in each vat, how much liquid is in each vat? (b) If you try to pour 234 gallons of liquid into buckets each with a capacity of 9 gallons, what is the minimum number of such buckets you need in order to hold these 234 gallons? Get the answer to both, and explain in each case whether you are using the partitive or the measurement interpretation of division.

3.5

The long division algorithm

Suppose we have to do the division problem 7864 ÷ 19. Up to this point, the only way we can do it is to look at all the multiples of 19 until we get a whole number q so that q × 19 ≤ 7864 but (q + 1) ×19 > 7864. Of course we could painstakingly go through all the multiples one by one until we hit one with the above property, but that would be dull bookkeeping rather than mathematics. Let us do better. First of all, we can ignore small multiples like 10×19 or even 100×19 because all we care about is getting a multiple of 19 that is close to 7864. Let us make an estimate: the 100th multiple of 19 is 1900, so the 400th multiple is 7600, which is close to 7864. Add ten more of 19 and we get: 7600 + 190 = 7790, which is even closer to 7864. A little experimentation shows that 7790+3×19 = 7847 < 7864 and 7790+(19×4) = 7866 > 7864. Because 7790 + (3 × 19) = (410 × 19) + (3 × 19) = 413 × 19 by the distributive law, we know 413 is the quotient. This method of ﬁnding the quotient is

should produce the correct answer of 195 and 1 in (38). by verifying that 195 × 3 = 585 so that adding 1 to it produces 586. our goal is to ﬁnd an eﬃcient algorithm that produces the quotient and remainder of a ÷ b.3 The Standard Algorithms
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clearly superior to the monotonous checking of all the multiples of 19 and deserves to be made more systematic. say a = 586 and b = 3. the correctness of (38) is easily checked. this line of thinking would lead us to the long division algorithm.) We know the conclusion we are supposed to draw from this: the mechanism described in (37) produces the quotient 195 and remainder 1. which is a beautiful and sophisticated method of ﬁnding the quotient and the remainder of a division-with-remainder. In other words. Without further ado. seemingly unrelated to multiplication or division in the usual sense we understand it. 586 = ( 195 × 3) + 1 (38)
The question is why ? Because this question can be easily misunderstood. According to the second form of the division-with-remainder.) Let us illustrate the algorithm we are after. after all. b. we bring the 6 down at each step of the long division in (37). With a little more work. this is the same as ﬁnding a q and an r so that a = qb + r and 0≤r<b
(See (35). The question here is not why (38) is correct. by something relatively simple. long division algorithm. The question is rather why the particular procedure adopted in (37). here is the usual schematic presentation of this algorithm for 586 ÷ 3: 1 9 5 3 ) 5 8 6 3 2 8 6 (37) 2 7 1 6 1 5 1 (Note that for reason of clarity of exposition. with b > 0. The failure to directly address this question
. let us explain it further. Given two whole numbers a.

. and the remainder 2 is really 200. how can we be using the division-with-remainder itself? Here is the main point: for “small” numbers. we interject a word of caution: there will be many computations of this type in this subsection.g. and 16 ÷ 3). this particular division is really 500 ÷ 3.. You can see from (37) that. Therefore: 586 = 500 + 86 = {([100 × 3] + 200)} + 86 = (100 × 3) + 286 .g. It is. 28 ÷ 3. So the ﬁrst step in (33) is actually a restatement of the division-with-remainder 500 = ( 100 × 3) + 200 . First. 28 ÷ 3. and then string the latter together in an artful way so as to get at the quotient and remainder of the division of the original large number. the division-with-remainder is used three times: 5 ÷ 3. in arriving at the purported quotient 195. we make a general comment about the long division algoithm n order to clarify our subsequent discussion. There is nothing wrong with the algorithm. the quotient and remainder of a division can be easily guessed at (e. and 16 ÷ 3. 586. it has already been stated above that it is one of the most beautiful pieces of elementary mathematics. (At this point. There is plenty that is wrong with the way elementary mathematics is taught.3 The Standard Algorithms
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in school mathematics and pre-service professional development materials is what makes the long division algorithm so notorious. Let us proceed to make amends by explaining the underlying reason why (37) leads to (38) simply and correctly. however. only a preliminary explanation. but since 5 stands for 500 in 586. so what the long division algorithm does is to break up the division of a large number (e. 5 ÷ 3. and we have to resist the temptation
It is a basic strategy in mathematics to try to break up a complicated task into a series of simpler tasks. we emphasize.12 Our task is to understand why this “stringing together” makes sense. The ﬁrst step in (37) looks like 5 ÷ 3.
12
. Now we are going to give a preliminary explanation of why the steps in (37) lead to the correct conclusion (38). Now you may ﬁnd the following fact puzzling: if we are trying to ﬁnd an algorithm which is more eﬃcient than the division-with-remainder itself in order to get at the quotient and the remainder. because we shall subsequently point out in what way it is unsatisfactory. although you can easily put up a number as large as you want) into a sequence of divisions of smaller numbers.

then we may also interpret n as the number of dollars in a stack when 3n dollars are divided equally into 3 stacks. and for this reason we want to keep the factor 3 intact all through these computations. Thus. The reason is that we want the ﬁnal outcome to be (195 × 3) + 1 as in (38). 586 = (100 × 3) + (90 × 3) + (5 × 3) + 1 = ([100 + 90 + 5] × 3) + 1 = ( 195 × 3) + 1 . or 90 × 3 in the succeeding sentence. where we have pointed out that multiplication and division undoes each other (at least in the case of no remainder). Suppose we have $586 consisting of 5 8 6 hundred-dollar bills ten-dollar bills one-dollar bills. we
. We next give an interpretation of (37) in terms of money. we begin the process of creating these 3 stacks by ﬁrst distributing the 5 hundred-dollar bills equally into these 3 stacks. It therefore follows that if division undoes multiplication. and the reason why this law must play a role is not entirely obvious. and there are 2 left over. Next.3 The Standard Algorithms
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of multiplying out thing like 100 × 3 above. the key reason underlying the multiplication algorithm is the distributive law.3. the long division algorithm must likewise make critical use of the distributive law. and the multiplicative algorithm depends critically on the distributive law. This corresponds to the ﬁrst step of (37). As we stressed in §3. 586 = (100 × 3) + 280 + 6 = (100 × 3) + ([90 × 3] + 10) + 6 = (100 × 3) + (90 × 3) + 16. In each stack we put in 1 hundred-dollar bill.
Although we are trying to ﬁnd out how many 3’s there are in 586. To this end. The last step is easy: 16 = ( 5 × 3) + 1 . It has to do with (35).) The second step in (37) is 28 ÷ 3. which is exactly (38). So. which as before is in reality 280 ÷ 3. We have of course used the distributive law. The division-with-remainder in this case reads: 280 = ( 90 × 3) + 10 . we can turn the original problem around: suppose there are n 3’s in 586 (with 0 or 1 or 2 left over).

with 1 one-dollar bill left over. and 1 is left over. To make sure that the basic facts of the long division algorithm and the preliminary explanation are understood. the original stack of $586 has been divided into three equal stacks each consisting of 1 hundred-dollar bill. we do another example: 1215 ÷ 35: 3 3 5 ) 1 2 1 1 0 5 1 6 1 4 2 Here is an abbreviated explanation: 1215 = = = = = = 1210 + 5 {(30 × 35) + 160} + 5 (30 × 35) + 165 (30 × 35) + ([4 × 35] + 25) ((30 + 4) × 35) + 25 ( 34 × 35) + 25 4 5 5 0 5 (39)
For an in-depth understanding of the long division algorithm. however. that the use of money is only an aid to understanding and is not to be confused with genuine mathematical understanding itself. the pre-
. Readers of this monograph cannot fail to realize. There are some people in mathematics education who consider the use of money to interpret the long divison algorithm as the height of conceptual understanding.3 The Standard Algorithms
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convert the 2 hundred-dollar bills into 20 ten-dollar bills. 9 ten-dollar bills. and we now have 16 one-dollar bills. and 5 one-dollar bills. This corresponds to the second step of (37). This is exactly what (38) says. so that (together with the original 8 ten-dollar bills already there) we now have 28 ten-dollar bills. Altogether then. Again. Observe that the preceding interpretation of (37) in terms of money is very similar in spirit to the intuitive approach to ﬁnding a quotient described at the beginning of this sub-section. Finally. We will in fact take up the mathematical explanation of long division next. we can distribute them equally into the three stacks with 5 one-dollar bills in each. we convert the 1 ten-dollar bill into 10 one-dollar bills. These can be distributed into these three stacks equally with 9 in each stack and 1 left over.

We repeat: the algorithm will go through each digit of 586. As we have emphasized throughout our discussion of algorithms.e. The long division algorithm is the most remarkable embodiment of these features among the algorithms we have studied. mechanically. because the algorithm will be repeating this step ever after. using as dividend the leftmost digit of the original dividend. So the ﬁrst division is 5 ÷ 3. We now give a mathematical explanation that is free of these defects. We start from the left. it lacks simplicity. Let us revisit (37). the number to be divided). one at a time.3 The Standard Algorithms
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ceding analysis falls short in two respects.e. for a change. 1 in (37) and the numbers 16 and 25 in (39)) which are critical to the understanding of the conversion of fractions to decimals in §4 of Chapter 4. The idea will be seen to be prefectly general. with absolutely no thought given to “breaking the dividend into parts” (as is taught in the schools). Look at the “dividend” 586 (i. every one of the standard algorithms gains eﬃciency and simplicity by ignoring place value and by performing the operations one digit at a time. each of which performs a division-withremainder The divisor will always be the original divisor (in this case it is 3).. First. and the dividend of the ﬁrst step is the ﬁrst digit of the original dividend (in this case it is the digit 5 of 586). There will be a sequence of steps.
. 1. Second. the numbers 2. in fact. It does not lead from a clear description of the algorithm straight to the desired conclusion about quotient and remainder.. The division-with-remainder gives 5 = ( 1 × 3) + 2 The next (second) step is the crucial one. So the only thing that needs to be speciﬁed in each step is the dividend. the explanation does not clearly expose the role played by the sequence of remainders (i. a clear description of the algorithm was never given. More formally: Step 1: perform the division-with-remainder. and we shall describe the long division algorithm precisely in this special case. The description of the dividend in the second step is this: Step 2: Multiply the remainder of the preceding step by 10 and add to it the next digit (to the right) in the original dividend.

If you are baﬄed by these equations. Thus the third step of the long division algorithm is: 16 = ( 5 × 3) + 1 Now (37) is entirely encoded remainders: 5 28 16 in the following three (simple) division-with= ( 1 × 3) + 2 = ( 9 × 3) + 1 = ( 5 × 3) + 1
(40)
You could not possibly fail to observe that the quotient 195 is clearly displayed in (40) — read vertically down the ﬁrst digits of the right sides — as well as the remainder 1 (the last term of the last equation). But of course one must keep in mind that (40) is true regardless of any such monetary interpretations. you can also read oﬀ the original dividend by going down the left sides of these equations and pick out the last digit of each number (in this case. so the dividend of the next step is 1 × 10 + 6 = 16. So the number in question is (2 × 10) + 8 = 28.
13
. and the third is the splitting of the 16 one-dollar bills into three equal stacks with also 1 left over. The second is the splitting of the 28 ten-dollar bills into three equal stacks with 1 left over. Now divide 28 by the same divisor 3: 28 = ( 9 × 3) + 1 We are now on automatic pilot: Step 3: repeat step 2. the next digit in the original dividend 586 is 6. The ﬁrst equation is a restatement of the splitting of the 5 hundred-dollar bills into three equal stacks with 2 left over. for example. You can. In the case of 586 ÷ 3. an algorithm is a sequence of mechanical procedures. (40)
Recall what was said at the beginning of §3. As we have emphasized. you get 586).13 With this in mind. Steps 1–3 explain how to generate these procedures as we go through the digits of the original dividend one-by-one.3 The Standard Algorithms
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In the present situation. let us hasten to point out that there is no mystery to them at all. relate them to the naive interpretation of (37) in terms of money in the following way. Though of slightly less interest. and the next digit of the original dividend 586 is 8. to the eﬀect that each standard algorithm would break a computation down to computations with single digit numbers. the remainder from the ﬁrst step is 2.

We now show how to generate (38) by the use of the long division algorithm as encoded in (40): 586 = (5 × 102 ) + (8 × 10) + 6 = {([3 × 1] + 2) × 102 } + (8 × 10) + 6 by the ﬁrst equation of (40). Obviously you have taken this for granted all along. Third.
Let us review what we have accomplished. because this is never mentioned in school mathematics. 1. which is (40). 1 of the long division algorithm (in the case of 586 ÷ 3) which will be of critical importance later in Chapter 4.3 The Standard Algorithms
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gives the procedures explicitly. the quotients 1. Now apply the last equation of (40) to get: 586 = = = = = (102 × 3) + ([9 × 10] × 3) + 16 (102 × 3) + ([9 × 10] × 3) + (5 × 3 + 1) (102 × 3) + ([9 × 10] × 3) + (5 × 3) + 1 {(102 + [9 × 10] + 5) × 3} + 1 (195 × 3) + 1 . 9. It is the fact that at each step of the algorithm ((37) or (40)). but which needs to be aired now. and 5 are always single digit numbers. Second. First we have veriﬁed (38) directly from a clearly stated digit-by-digit description of the long division algorithm. This gives an explicit support to the interpretation about splitting the 5 hundred-dollar bills into three equal stacks with 2 left over. the simplicity of the long
. The last line corresponds to the splitting of the 28 ten-dollar bills into tghree equal stacks with 1 left over. Nevertheless. there is a point which has been purposely suppressed thus far in order to get across the main thrust of the argument as clearly as possible. But to continue: 586 = = = = (3 × 102 ) + (2 × 102 ) + (8 × 10) + 6 (102 × 3) + (20 × 10) + (8 × 10) + 6 (102 × 3) + (28 × 10) + 6 (102 × 3) + ([3 × 9] + 1) × 10 + 6
by the second equation of (37). (40) exhibits the sequence of remainders 2.

It is straightforward to see from (40) — and this is another reason why the explicit description of the algorithm in Steps 1–3 is critical — that because each remainder is by deﬁnition < 3. 8 by 9 and 6 by 5. the ﬁrst form of the Division-with-Remainder in §3. Therefore the dividend at each step of the long division algorithm is always < 30. so that the next dividend (i. 5 by 1. 8.. we note that this reasoning is valid in general. and 6 at the beginning. a digit of the dividend could be replaced by 0) so that this “single-digit” fact should be carefully veriﬁed. The usual schematic display of the long division is as follows: 3 1 ) 5 3 2 2 9 5 5 7 8 6 7 1 8 7 1 6 1 5 1 7 1 5 2 1 2 1 0
The precise description of the algorithm in accordance with Steps 1–3 is then:
. then its quotient when divided by 3 must be a single digit. where the “dividend” 58671 has been chosen on purpose to have the three digits 5.e. (Cf. and the number 16 after the second) is ≤ (20 + a single digit number) according to Step 2. As usual.3 The Standard Algorithms
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division algorithm depends on “replacing” each digit of the original dividend by another digit (e. of course on other occasions. This we proceed to do.g. the number 28 after the ﬁrst step.. To this end we shall compare (37) with the division 58671 ÷ 3. But if a number is < 30.4). We want to drive home the point that the long division algorithm is strictly a digit-by-digit procedure without regard to place value. exactly as claimed. it is ≤ 2.

Here then is 11546 ÷ 19 :
. The point we wish to emphasize is that. in order to illustrate more clearly why it is unnecessary to worry about “breaking the dividend into parts”. the third step in (41) is actually 1600 ÷ 3 and the corresponding division algorithm is then 1600 = (500 × 3)100 if place value is taken into consideration. which is very diﬀerent from the 6 in 58671. carried out in accordance with Steps 1–3. Contrast this with the ﬁrst explanation given of (37). which is laden with place value interpretations. This is the point we made earlier about the lack of simplicity in the latter explanation. In other words. and the quotient 19557 can be read oﬀ by going down the ﬁrst digits of the right sides. As a reminder: observe once again that in (41). Moreover. let us concentrate on the third step in both. They are identical. which is in the hundreds digit. Needless to say. The 6 in 586 is in the ones digit. as far as the long division algorithm itself is concerned. the dividend 58671 can be read oﬀ by going down the last digits of the left sides. place value is irrelevant.3 The Standard Algorithms
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5 28 16 17 21
= = = = =
( ( ( ( (
1 9 5 5 7
× 3) + × 3) + × 3) + × 3) + × 3) +
2 1 1 2 0
(41)
Now compare the ﬁrst three steps of both (40) and (41). For deﬁniteness of discussion. Let us give an example of a long division where the divisor has more than one digit. the explanation of why (41) leads to 58671 = 3 × 19557 — along the line of the argument leading from (40) to (38) — is squarely based on (41) and nothing else. the remainder (0) appears in the last equation.

e.
4 The Number Line and the Four Operations Revisited
We introduced the number line in §2 as an “inﬁnite ruler” with the whole numbers identiﬁed with a set of equally spaced points (often referred to as “markers”) to the right of a point designated as 0..) Exercise 3.4 The Number Line and the Four Operations Revisited
90
Exercise 3. we shall be concerned exclusively with the part of the number line to the right of 0. Then write out the procedural description of this long division along the line of (40). Exercise 3. (In other words. and use it to explain why your result is correct. and this terminology has to be understood in the following sense. A whole number n is also identiﬁed with the length of the line segment [0. we shall always refer to the number line. and the number 1 is sometimes referred to as the unit. n] from 0 to n. The positions of the whole numbers depend completely on the choice of 0 and 1. The segment [0. Exercise 3. Until the end of Chapter 4. if the quotient and remainder of your original long division are respectively q and r. for example). i. use your procedural description to show directly that 10192 = (q × 8) + r. the positions of the other whole numbers are likewise ﬁxed.27 Compute 10192 ÷ 8 using the long division algorithm. we always assume that a unit segment has been chosen on the given straight line so that the whole numbers are ﬁxed on the line. There will be occasions when we see ﬁt to change the unit. we know you can multiply 3 × 19557 to get 19557. but we’d prefer that you learn how to use a sequence of division-with-remainders such as (41) to explain the long division algorithm.30 Do the long division of 50009 ÷ 67 to ﬁnd the quotient and remainder. In each discussion. Be sure you write down every step of the procedural description (as in (42). and use these to show why your quotient and remainder are correct. in which case there will be a diﬀerent number
.28 Do the same for 21850 ÷ 43.29 Use (41) to derive the fact that 58671 = 3 × 19557 . Once these two numbers have been ﬁxed. In the following. 1] is called the unit segment. It is in this sense that the number line is ﬁxed. describe the algorithm as a sequence of division-with-remainders in accordance with Steps 1–3.

) In any case. perhaps one can describe it as the stone-age version of the calculator. (For those who do not know what a “slide rule” is. m + n = the length of the segment obtained by concatenating the segments [0. The four arithmetic operations have also been interpreted geometrically. as indicated below.n]
This way of adding numbers is exactly the principle underlying the slide rules of yesteryear.m]
[0. For addition. n] Geometrically. Subtraction is next. Suppose we have two number lines. especially with respect to the number line. then we saw in §3.2 that n − m = the length of the segment obtained when a segment of length m is removed from one end of a segment of length n In picture:
'
(44)
n−m
E '
m n
'
E E
We wish to go into some of the ﬁne points of the addition and subtraction of whole numbers.
. one can do many activities with (43) until this geometric way of adding numbers become second nature. n are whole numbers and m < n. If m. we have that for any two whole numbers m and n.4 The Number Line and the Four Operations Revisited
91
line to deal with. m] and [0. we have: m+n (43)
'
E
[0. Such an occasion will come up soon enough.

and concatenate them as shown:
'
1+2
E
[0.4 The Number Line and the Four Operations Revisited
92
0 0
1
2 1
3
4 2
What do you say to a student if she tells you that she gets 1 + 2 = 2 in the following way: She takes [0. 2] from the upper number line. 1] from the lower number line and [0. Consider for example the following equations: 9−2 = 1 8 + 16 = 2 19 + 17 = 3 (45)
Although every equation in (45) is wrong according to the arithmetic of whole numbers as we know it. and therefore are done with respect to a ﬁxed unit segment. and subsequently got the two unit segments mixed up. including (43) and (44).2]
According to the lower number line. it is not as absurd as it appears. This brings up a fundamental issue in the arithmetic operations on whole numbers concerning the importance of having the same unit as a ﬁxed reference. So what she did wrong was not to realize that she had changed her unit segment in going from the upper number line to the lower number line. and this is how she gets 2 for her answer. you would have to recall for her the fact that all geometric representations of operations on whole numbers.1]
[0. the resulting segment has length 2. The question is: what is wrong? In order to explain to her what the mistake is. What (45) wants to say is the following:
. are done on one number line.

the ﬁrst interpretation of 2 + 3 = 5 is based on taking 1 to be “one apple”. Suppose a ring R is isomorphic to the integers Z. Now suppose we decide that 1 is the area of a ﬁxed square. In this context. each addition or subtraction must refer to the same unit. Recall the convention that once we decide on a ﬁxed segment as the unit segment. So the number “1” will henceforth refer to this unit area. However. Thus. the second on “one cup of coﬀee”. we come to understand (43) as the interpretation of 1 as a ﬁxed unit length. then the area of the unit square (= the square with each side of length equal to that of the unit segment) deﬁnes the unit area. The concept of a whole number is an abstract one: for example. and the third on “one square inch”. Then although both ¯ + ¯ and 2 + 5 make perfect 2 5 sense. Then the number 3 is no longer a concatenation of three unit segments but rather the
. we cannot perform the addition ¯ + 5 or 2 + ¯ 2 5. so that each addition is nothing but combining unit lengths of segments. and suppose under the isomorphism n ∈ R ¯ corresponds to n ∈ Z.4 The Number Line and the Four Operations Revisited
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9 days − 2 days 8 months + 16 months 19 eggs + 17 eggs
= 1 week = 2 years = 3 dozen eggs
The point of (45) is to underline the implicit or explicit role played by the unit in any addition or subtraction of numbers. it can never be interpreted to mean 2 apples and 3 cups of coﬀee are the same as 5 square inches It is for the same reason — changing the unit in the middle of addition — that the student’s reasoning of 1 + 2 = 2 is wrong.
This discussion of units assumes special importance when we discuss length and area.
In a mathematical context. the equation 2 + 3 = 5 could mean any of the following among numerous other possibilities: 2 apples and 3 apples are the same as 5 apples 2 cups of coﬀee and 3 cups of coﬀee are the same as 5 cups of coﬀee 2 square inches and 3 square inches are the same as 5 square inches Whatever the interpretation of the abstract operations. what we are saying is this.

5 would be the area of the following rectangle:
If there is no fear of confusion. in this particular context. 2 + 3 would be a concatenation of the following two rectangles rather than a concatenation of the segments [0. we shall agree to interpret a whole number n in this context as the area of a rectangle whose width is a unit segment and whose length is the concatenation of n unit segments. 2 × 3 is the area of the rectangle:
(47)
Therefore 2 × 3 = 6 means that the area of the the rectangle of (47) is the same as the area of
because the latter is exactly what 6 stands for in this context. For convenience.4 The Number Line and the Four Operations Revisited
94
total combined area of three unit squares. we would simply say this is a rectangle of width 1 and length 5. 3] as stated in (43):
(46)
What makes this discussion particularly relevant is that we have interpreted the multiplication of whole numbers in §2 as area.g. e. the two numbers are
. the equality 2 × 3 = 6 means if we count the number of unit squares on the left and count the number of unit squares on the right. 2] and [0. For instance. to remind you of the meaning of the equal sign “=” as explained at the beginning of §2. (Or. Again..

The ﬁrst one is that declaring the area of the unit square (recall: this is the square whose side has length 1) to be 1 is nothing more than a convention. Thus declaring the area of the unit square to be anything other than 1 would serve no purpose other than messing up otherwise simple formulas. we all agree to set the area of the unit square to be 1. and not (5 × 7) (which would be the case if each unit square has area equal to 1). Look at the area of the rectangle with width 5 and length 7. For this same reason. “contained by” means “having for its sides”. and “the line m” means “the line segment of length m”). This rectangle is paved (tiled) by 5 × 7 unit squares:
Therefore the area of the rectangle is 2 × (5 × 7). We shall be discussing the area interpretation of numbers extensively in Chapter 2. A second comment is that the representation of the product of whole numbers as the area of the corresponding rectangle has a long history behind it. Each time Euclid wanted to express that idea. a product of three numbers. For this reason. say. More generally. Until the time of Descartes (1596-1650). for example. we give one possible interpretation of 5 + (2 × 3) = 11: it means that the area obtained by combining the rectangle of (46) and the rectangle of (47) is the same as the area of the rectangle whose width is 1 and whose length is 11. then its area would be 2mn intead of the usual mn. there was never any mention of multiplying two numbers m and n.4 The Number Line and the Four Operations Revisited
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the same.).C. Euclid’s Elements (circa 300 B. if a rectangle has width m and length n.E. as follows. we could declare its area to be any number. In the most inﬂuential mathematics textbook of all time. In fact. such as 12 × 7 × 9. What could go wrong then? The area formula for the rectangle would be messed up. had to be interpreted as volume
. this was the only way to understand the multiplication of numbers. 2. he would say: “the rectangle contained by the line m and the line n” (translation: “the rectangle” in Euclid means “the area of the rectangle”.
Two extra comments would help to clarify the circle of ideas in connection with the representation of 1 as the area of the unit square.) To drive home this point.

For this monograph. Schematically. if m and n are whole numbers. say m = 4 and n = 3. a (good) college course on number systems would develop all number concepts in an abstract setting without reference to geometry. The purely algebraic approach to multiplication will be discussed in §7. in terms of such a choice of the unit 1. The product 3×4. Nowadays. one can think of 4 as a bag of 4 potatoes. the number 3 is the point on the line represented by the following 3 groups of objects: • • • • • • • • • • • •
In general. then mn may be interpreted as the number m on the number line whose unit 1 is taken to be (the magnitude or size represented by) the number n. can be interpreted as the number 3 on a number line whose unit 1 is taken to be (the magnitude or size represented by) the number 4. So m in this context is the point on the line represented by the following m groups of objects:
n objects
n objects
···
n objects
m
For a later need.4 The Number Line and the Four Operations Revisited
96
and therefore the product of four or more numbers was almost never considered until Descartes pointed out that multiplication can also be regarded as an abstract concept independent of geometry. the geometric interpretation of multiplication not only is convenient for our purpose. a bag of 4 marbles. and we shall re-interpret 4 × 3 as the point 4 on a number line with a new unit. but has the added advantage in that it is suﬃciently similar to the common manipulative of Base Ten Blocks to make a beginner feel at ease. one that was mentioned in §2. for example. a box of 4 crayons. So we start with a number line:
. let us give m and n explicit values.
At this point.2 of Chapter 2 and also §3 of Chapter 5. To facilitate the discussion. a car full of 4 people. etc. however. we are in a position to give another interpretation of the multiplication of whole numbers. it would be advantageous to formalize this procedure.

say a = qb for some whole number q. multiplication and division undoes each other. Suppose. namely. etc. If one gives a little thought to what m ÷ n = k could mean regardless of
. Activity: Suppose a fourth grader understands 24 ÷ 3 = 8 only in terms of the measurement interpretation and the partitive interpretation of division. and 3m would be just m for any whole number m.3 of Chapter 2. To avoid confusion. let us assume for now that a is a multiple of b. 2 is right under 6. we also say multiplication and division are inverse operations to express this fact. we shall distinguish the new number markings from the original one by a bar and place them underneath the line. Then we write a÷b=q for a = qb. division. b. 0 1 2 3 1 4 5 6 2 7 8 9 3 10 11 12 4
This idea of using a new unit to re-interpret the multiplication of numbers provides an alternative way to understand the multiplication of fractions. This is the place to tie up a loose end mentioned in the discussion of the division algorithm. So 1 is right under 3. Finally. with b always assumed to be nonzero. (See (36) and the discussion surrounding it. (48)
In intuitive language. 4 × 3 is four copies of the new unit and is therefore 4 in the new number line. In particular.4 The Number Line and the Four Operations Revisited
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0
1
2
3
4
5
6
7
8
9
10
11
12
Now introduce new markers on the same line. why one cannot divide by 0. say n ÷ 0 = 3. Explain to her why 24 ÷ 3 = 8 is the same as 24 = 8 × 3. where the new unit is 3. For whole numbers a.) Sometimes. as we shall see in §7. division by zero makes sense for a particular nonzero whole number n.

We have therefore shown that the value of 0 ÷ 0 is ambiguous. the partition of the n objects cannot be done. The fact that division undoes multiplication (always understood in the sense of (36) or (48)) leads to a geometric interpretation of division that has already been mentioned at the end of §3.. What about 0 ÷ 0 ? We now run the preceding argument backwards. So n ÷ 0 cannot be equal to any whole number if n is nonzero. one would likely conclude (with (48) as guide) that it means m = nk. it could be 1. n. so 0 ÷ 0 = 2 . which is another way of saying it cannot be deﬁned. But it is also true that 0 = 0 × 2 . But if each part has no object. In other words. Then it is the number of parts when n objects are partitioned into diﬀerent parts so that each part has exactly 0 objects. i. By the partitive interpretation. and k. So again. this interpretation has no meaning either. n ÷ 0 would mean the number of objects in a part when n objects are partitioned into 0 equal parts (see §3. this has no meaning. the same argument applies. Such being the case. Assuming as always that a is a multiple of b. It may be instructive to follow literally the partitive and measurement interpretations of division to see why n ÷ 0 cannot be deﬁned for a nonzero whole number n. it remains to point out that al-
. i. Suppose it were deﬁnable. it is also undeﬁnable. n. Because we cannot partition anything into 0 equal parts. If 3 is replaced by any other whole number. m = nk should mean the same thing as m ÷ n = k for all m. We have therefore shown that division by 0 cannot be deﬁned. n.4). or 2.e.e. one would require that (48) makes sense for all m. k may be. or in fact any whole number by the same argument. Now suppose n ÷ 0 were meaningful in the measurement sense. then a ÷ b is the other side of a rectangle whose area is a and one of whose sides is equal to b:
a÷b
a b
(49)
To conclude this discussion of division. This shows that 0 ÷ 0 cannot be given a deﬁnite value.. Thus knowing 0 = 0 × 1 means 0 ÷ 0 = 1 . which contradicts our assumption that n is nonzero.4 The Number Line and the Four Operations Revisited
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what m.4. n ÷ 0 = 3 would be the same as saying n = 0 × 3 = 0.

it will be seen when we come to §8 of Chapter 2 that it is in fact no restriction at all once fractions are at our disposal. let us do it ﬁrst. we have never paused to ask what a whole number is. This generates the third marker. Before you ask why we bother. The equivalence of division and multiplication as described in (48) will be seen to be the key to the understanding of division in general. Exercise 4. mark the new position of the right endpoint of the unit segment. Take a straight line and mark oﬀ a point as 0 (zero). We are going to change our viewpoint here and use a set of markers on a line to deﬁne the whole numbers. Now we will formalize the introduction of whole numbers by adopting the following deﬁnition. and subsequently make them correspond to a collection of equally spaced markers (points) on a line to the right of a point denoted by 0. This generates a sequence of equally spaced markers to the right of 0. So we start afresh by imposing a set of markers on a line. there is no mention of whole numbers. thereby generating the ﬁrst marker.2 If a rectangle has area 98 and one side equals 14. Then ﬁx a segment to the right of 0 and call it the unit segment.5 What Is a Number?
99
though the restriction that a is a multiple of b imposed here seems too severe.1 Use the idea of introducing a new unit to represent a whole number c to re-interpret the distributive law in the form of (a + b)c = ac + bc. Now slide the unit segment to the right again until its left endpoint rests on the second marker. Mark the right endpoint of this segment on the line. Exercise 4. etc. and mark the right endpoint of the unit segment in its new position. what is the other side? If the area is 1431 and one side is 27? And if the area is 7797 and one side is 113?
5 What Is a Number?
Thus far. Notice that up to this point. We took this concept for granted from the beginning.
. thereby generating the second marker. Slide the unit segment to the right until its left endpoint is at the ﬁrst marker.

baﬄed the human race for over two thousand years before it was ﬁnally pinned down in the late nineteenth century. See Chapters 2. A number is by deﬁnition any point on the number line. the one after that is 2. 4 and 5. don’t we know what a whole number is? Let us deﬁne the number 5. We also know “ﬁve chairs”. we are not by any means trying to attempt a retrograde revision of our knowledge of the whole numbers. In other words. Thus a whole number is now something very concrete and explicit: it is among the markers on the number line which were carefully constructed above. so that. or ﬁve of anything we see or touch because we can count.5 What Is a Number?
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Deﬁnition. This deﬁnition of numbers is not ideal. the next one (to the right of 0) is 1. This line with the whole numbers on it is called the number line. Do not be discouraged. for instance. we should be able to explain it in terms of these points on the number line. What we need for elemetary mathematics is fortunately nothing very sohpisticated. what we have deﬁned is at least consistent with everything we have done up to this point. but rather a precise deﬁnition analogous to the deﬁnition (say) of a triangle as three noncollinear points together with the line segments joining them. because the general concept of a number. we shall agree to change our point of view and base our reasoning with whole numbers on this deﬁnition alone. for instance.. Only a small portion of those points. we will not scrutinize every point on the number line. we are saying that. for the work we do from now on. Fractions or rational numbers. You must be muttering to yourself at this point and wondering what is happening here. no revision of anything we have done is necessary. just the whole numbers and some other numbers which come out of whole numbers in a rather simple-minded fashion. We know “ﬁve ﬁngers”. Rather. without reference to any concrete object? So you see that it is diﬃcult. starting with the initial number 0. Note that what is required here is not a description of our intuitive feelings about “5”. but it serves our pedagogical needs admirably.
. As far as whole numbers are concerned. After all. or “ﬁve apples”. however. Because we have been using these markers all along in our work. etc. But ﬁve itself. in the sense that it is accessible and it lends itself to a reasonable treatment of rational numbers and decimals. A whole number is one of the markers on the line. in the sense deﬁned above as a point on the number line. and we continue the naming of the markers in the same way we did the counting of the whole numbers in §1. Whatever we do in the future about whole numbers. In this sense.

5 What Is a Number?
101
A precise deﬁnition of whole numbers is not strictly necessary if all we ever do in mathematics is to stay within the realm of whole numbers. most (or perhaps all) of the school textbooks and professional development materials would have you believe that whole numbers are simple. At the moment. then our chances of success in teaching them fractions would be immeasurably increased. We want to change this dismal scenario by adopting the down-to-earth approach of giving direct answers to direct questions. so much so that you can ﬁnd references to it in the comic strips of Peanuts and FoxTrot. For example. but fractions are a completely diﬀerent breed of animals. But we cannot stay with whole numbers forever. We will deﬁne fractions. The answer is that in order for children to understand fractions. we can communicate the essence of it by putting up one hand with the ﬁngers outstretched. or any concept we ever take up. The generic non-learning of fractions among children has become part of our national folklore. The minute you as a teacher or your students buy
. If we can convince them that fractions are nothing more than an natural extension of the whole numbers. that should be enough to communicate any kind of “ﬁveness” needed for conceptual understanding. because the next topic is fractions. This is the advantage of whole numbers: each has (at least in principle) a concrete manifestation such as outstretched ﬁngers that almost renders abstract considerations about whole numbers unnecessary in elementary school. What concrete image can one conjure 119 13 in connection with or ? Children need answers to this question in 7 872 their quest for knowledge because they need something to anchor the many concepts related to fractions in the same way a handful of ﬁngers can anchor any discussion about 5. even if we cannot deﬁne precisely what 5 is. Whole numbers are taught one way. school mathematics in our country has contrived to never answer this question. Learning is a gradual process ﬁrmly rooted in prior experiences. fractions cannot be suddenly presented to them out of the blue. Now you would want to know why not just deﬁne fractions abstractly and leave a nice subject like whole numbers alone. Amazingly. There is no continuity from one to the other. The results are entirely predictable: when adults abrogate their basic responsibilities. and fractions in a completely diﬀerent way. the ﬁrst victims are the children. decimals.

it is highly unlikely that they think about “how soon before I can breathe in and how many strides I should take before then” each time they run to catch a bus. the real numbers. it is not possible to only oﬀer a precise deﬁnition of fractions and leave the whole numbers unattended. (46)–(47) and (49) are part of this groundwork. He would tell you that he calculates exactly how many strides he takes with each breathe-in.5 What Is a Number?
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into this view of numbers. All you need to do is to understand decimals and fractions and all the rest. this is where the number line comes in: we already have the whole numbers there. doing what is natural is simply not enough. our mathematics education is already in trouble because this means you have bought into mathematical misinformation. if you talk to an Olympic sprinter. In fact. whole numbers are on an equal footing with fractions. and where the next spot is before he can take another breath again. Yet. how many he takes with each breathe-out. all of them would have been hunted down by the predators on the African Savannas and we wouldn’t be here to talk about fractions. you need not ﬁxate on the number line every time you count oranges in the supermarket. for instance. Let us consider the philosophical question of why something as natural as a whole number should be made into something as cold and formal as “a point on the number line”. We must begin with a deﬁnition of the whole numbers that will naturally lead to fractions. had our ancestors been less good at it. and the next step is to single out the fractions on this line. Whatever the Olympic sprinters do in a race. In particular. Now it is in the nature of human aﬀairs that each time we try to achieve excellence in any endeavor. In fact. So please rise to the occasion when there is a need to regard a whole number as a point on the number line and be willing to work with this concept.
. then you have already won half the battle. This is about as natural an activity as we are going to get. The answer lies in the fact that we are trying to deepen our understanding of the whole numbers in order to lay the groundwork for working with fractions. If you can accept this reality. They are part of the same family. exactly how far from the starting block before he can take his ﬁrst breath. For this reason. Doesn’t this unnnatural and calculated approach to running remind you of looking at a whole number as a point on the number line? But let us not lose our perspective. In the same way. what you hear from him about running would strike you as extremely unnnatural if not downright unreal. From the point view of mathematics. Take running. the geometric interpretations (43)– (44). and be good teachers.